Calculus 1 Study Notes: Rates of Change, Limits, and Continuity

Average Rates of Change and Secant Lines

  • Definition of Average Rate of Change (AROC):

    • The average rate of change of a function f(x)f(x) over a closed interval [a,b][a, b] measures the average amount by which the output changes per unit change in the input.

    • Formally, the average rate of change is given by the formula:     AROC=f(b)−f(a)b−a\text{AROC} = \frac{f(b) - f(a)}{b - a}

    • Geometrically, the average rate of change corresponds to the slope of the secant line passing through the points (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)) on the graph of the function.

  • Algebraic Calculations of Average Rates of Change:

    • Polynomial Function Example:

    • For the function g(s)=(s2+1)3g(s) = (s^2 + 1)^3 over the interval from s=0s = 0 to s=3s = 3:

    • Evaluate at endpoints:       g(0)=(02+1)3=13=1g(0) = (0^2 + 1)^3 = 1^3 = 1       g(3)=(32+1)3=(9+1)3=103=1000g(3) = (3^2 + 1)^3 = (9 + 1)^3 = 10^3 = 1000

    • Compute the difference quotient:       AROC=g(3)−g(0)3−0=1000−13=9993=333\text{AROC} = \frac{g(3) - g(0)}{3 - 0} = \frac{1000 - 1}{3} = \frac{999}{3} = 333

    • Algebraic Rational and Radical Function Example:

    • For the function t(y)=y+2−yyt(y) = \frac{y + 2 - \sqrt{y}}{\sqrt{y}} over the interval from y=4y = 4 to y=16y = 16:

    • Evaluate at endpoints:       t(4)=4+2−44=6−22=42=2t(4) = \frac{4 + 2 - \sqrt{4}}{\sqrt{4}} = \frac{6 - 2}{2} = \frac{4}{2} = 2       t(16)=16+2−1616=18−44=144=72t(16) = \frac{16 + 2 - \sqrt{16}}{\sqrt{16}} = \frac{18 - 4}{4} = \frac{14}{4} = \frac{7}{2}

    • Compute the average rate of change:       AROC=t(16)−t(4)16−4=72−212=3212=324=18\text{AROC} = \frac{t(16) - t(4)}{16 - 4} = \frac{\frac{7}{2} - 2}{12} = \frac{\frac{3}{2}}{12} = \frac{3}{24} = \frac{1}{8}

  • Graphical Approximations of Secant Line Slopes:

    • Parabolic Function:

    • For a parabolic graph evaluated on the interval −1≤x≤2-1 \le x \le 2:

    

Parabola Secant Line
* Identifying coordinates from the grid: (−1,0)(-1, 0) and (2,1.5)(2, 1.5).
* Calculating the slope of the secant line:

      Slope=1.5−02−(−1)=1.53=12\text{Slope} = \frac{1.5 - 0}{2 - (-1)} = \frac{1.5}{3} = \frac{1}{2}

  • Exponential Function:

    • For an exponential curve evaluated on the interval −1≤x≤3-1 \le x \le 3:

    

Exponential Secant Line
* Identifying points from the grid: (−1,0)(-1, 0) and (3,19)(3, 19) or (3,20)(3, 20).
* Calculating the secant line slope:

      Slope=20−13−(−1)=194\text{Slope} = \frac{20 - 1}{3 - (-1)} = \frac{19}{4}

  • Equations of Secant Lines in Slope-Intercept Form:

    • Cubic Function Example:

    • Function: y=h(t)=t3y = h(t) = t^3 over the interval [−1,1][-1, 1].

    • Evaluate points: h(−1)=(−1)3=−1h(-1) = (-1)^3 = -1 and h(1)=13=1h(1) = 1^3 = 1

    • Slope:       m=h(1)−h(−1)1−(−1)=1−(−1)1+1=22=1m = \frac{h(1) - h(-1)}{1 - (-1)} = \frac{1 - (-1)}{1 + 1} = \frac{2}{2} = 1

    • Line equation using point (1,1)(1, 1):       y−1=1(t−1)  ⟹  y=ty - 1 = 1(t - 1) \implies y = t

    • Quadratic Product Example:

    • Function: y=j(z)=(z−1)(z−2)y = j(z) = (z - 1)(z - 2) over the interval [1,2][1, 2].

    • Evaluate points: j(1)=(0)(−1)=0j(1) = (0)(-1) = 0 and j(2)=(1)(0)=0j(2) = (1)(0) = 0

    • Slope:       m=j(2)−j(1)2−1=0−01=0m = \frac{j(2) - j(1)}{2 - 1} = \frac{0 - 0}{1} = 0

    • Line equation:       y=0y = 0

Estimating Instantaneous Rates of Change

  • Method of Shrinking Intervals:

    • The Instantaneous Rate of Change (IROC) of a function f(x)f(x) at a point x=ax = a represents the slope of the tangent line at that exact point.

    • IROC is estimated by calculating average rates of change over progressively smaller intervals [x,x+h][x, x + h], as hh approaches 00 (h=1,0.1,0.01,0.001,0.0001h = 1, 0.1, 0.01, 0.001, 0.0001).

  • Linear Function Analysis:

    • For f(x)=7x+3f(x) = 7x + 3 at x=9x = 9:

    • Interval h=1h = 1 ([9,10][9, 10]):       AROC=f(10)−f(9)10−9=(7(10)+3)−(7(9)+3)1=73−661=7\text{AROC} = \frac{f(10) - f(9)}{10 - 9} = \frac{(7(10) + 3) - (7(9) + 3)}{1} = \frac{73 - 66}{1} = 7

    • Interval h=0.1h = 0.1 ([9,9.1][9, 9.1]):       AROC=f(9.1)−f(9)0.1=(7(9.1)+3)−660.1=66.7−660.1=0.70.1=7\text{AROC} = \frac{f(9.1) - f(9)}{0.1} = \frac{(7(9.1) + 3) - 66}{0.1} = \frac{66.7 - 66}{0.1} = \frac{0.7}{0.1} = 7

    • Interval h=0.01h = 0.01 ([9,9.01][9, 9.01]):       AROC=f(9.01)−f(9)0.01=0.070.01=7\text{AROC} = \frac{f(9.01) - f(9)}{0.01} = \frac{0.07}{0.01} = 7

    • Conclusion: For any linear function f(x)=mx+bf(x) = mx + b, the average rate of change over any interval equals the constant slope mm, so the estimated IROC at x=9x = 9 is strictly 77

  • Quadratic Function Analysis:

    • For f(x)=x2−9f(x) = x^2 - 9 at x=3x = 3:

    • Interval h=1h = 1 ([3,4][3, 4]):       AROC=f(4)−f(3)4−3=(42−9)−(32−9)1=7−01=7\text{AROC} = \frac{f(4) - f(3)}{4 - 3} = \frac{(4^2 - 9) - (3^2 - 9)}{1} = \frac{7 - 0}{1} = 7

    • Interval h=0.1h = 0.1 ([3,3.1][3, 3.1]):       AROC=f(3.1)−f(3)3.1−3=(3.12−9)−00.1=9.61−90.1=0.610.1=6.1\text{AROC} = \frac{f(3.1) - f(3)}{3.1 - 3} = \frac{(3.1^2 - 9) - 0}{0.1} = \frac{9.61 - 9}{0.1} = \frac{0.61}{0.1} = 6.1

    • Interval h=0.01h = 0.01 ([3,3.01][3, 3.01]):       AROC=f(3.01)−f(3)3.01−3=(3.012−9)−00.01=9.0601−90.01=0.06010.01=6.01\text{AROC} = \frac{f(3.01) - f(3)}{3.01 - 3} = \frac{(3.01^2 - 9) - 0}{0.01} = \frac{9.0601 - 9}{0.01} = \frac{0.0601}{0.01} = 6.01

    • Interval h=0.001h = 0.001 ([3,3.001][3, 3.001]):       AROC=f(3.001)−f(3)0.001=0.0060010.001=6.001\text{AROC} = \frac{f(3.001) - f(3)}{0.001} = \frac{0.006001}{0.001} = 6.001

    • Conclusion: As h→0h \to 0, the average rate of change approaches 66. Therefore, the estimated IROC at x=3x = 3 is IROC=6\text{IROC} = 6

Difference Quotients and Derivatives

  • General Difference Quotient Formula:

    • The average rate of change of a position function x(t)x(t) between time tt and time t+ht + h is expressed by the difference quotient:     Average Velocity=x(t+h)−x(t)h\text{Average Velocity} = \frac{x(t + h) - x(t)}{h}

    • Simplifying the difference quotient algebraically so that it is continuous at h=0h = 0 allows direct evaluation of instantaneous velocity v(t)v(t) by substituting h=0h = 0

  • Reciprocal Function Position:

    • Position function: x(t)=1t ftx(t) = \frac{1}{t}\,\text{ft} for t>0 st > 0\,\text{s}.

    • Difference quotient derivation:     x(t+h)−x(t)h=1t+h−1th=t−(t+h)t(t+h)h=−ht(t+h)h=−1t(t+h)\frac{x(t + h) - x(t)}{h} = \frac{\frac{1}{t + h} - \frac{1}{t}}{h} = \frac{\frac{t - (t + h)}{t(t + h)}}{h} = \frac{\frac{-h}{t(t + h)}}{h} = \frac{-1}{t(t + h)}

    • Instantaneous velocity v(t)v(t) at h=0h = 0:     v(t)=−1t(t+0)=−1t2 ft/sv(t) = \frac{-1}{t(t + 0)} = \frac{-1}{t^2}\,\text{ft/s}

    • Instantaneous acceleration a(t)a(t):     a(t)=v′(t)=2t3 ft/s2a(t) = v'(t) = \frac{2}{t^3}\,\text{ft/s}^2

  • Inverse Square Position Function:

    • Position function: x(t)=1t2 ftx(t) = \frac{1}{t^2}\,\text{ft} for t>0 st > 0\,\text{s}.

    • Difference quotient derivation:     x(t+h)−x(t)h=1(t+h)2−1t2h=t2−(t+h)2t2(t+h)2h=t2−(t2+2th+h2)ht2(t+h)2=−2th−h2ht2(t+h)2=−2t−ht2(t+h)2\frac{x(t + h) - x(t)}{h} = \frac{\frac{1}{(t + h)^2} - \frac{1}{t^2}}{h} = \frac{\frac{t^2 - (t + h)^2}{t^2(t + h)^2}}{h} = \frac{t^2 - (t^2 + 2th + h^2)}{h t^2 (t + h)^2} = \frac{-2th - h^2}{h t^2 (t + h)^2} = \frac{-2t - h}{t^2 (t + h)^2}

    • Instantaneous velocity v(t)v(t) at h=0h = 0:     v(t)=−2t−0t2(t+0)2=−2tt4=−2t3 ft/sv(t) = \frac{-2t - 0}{t^2(t + 0)^2} = \frac{-2t}{t^4} = \frac{-2}{t^3}\,\text{ft/s}

Physical and Applied Applications of Rates of Change

  • Circular Puddle Expansion:

    • Radius equation: r(t)=t cmr(t) = \sqrt{t}\,\text{cm}, where tt is time in seconds.

    • Area equation as a function of tt:     A(t)=π[r(t)]2=π(t)2=πt cm2A(t) = \pi [r(t)]^2 = \pi (\sqrt{t})^2 = \pi t\,\text{cm}^2

    • AROC of Area with respect to Time (t=0t = 0 to t=16 st = 16\,\text{s}):     AROC=A(16)−A(0)16−0=16π−016=π cm2/s\text{AROC} = \frac{A(16) - A(0)}{16 - 0} = \frac{16\pi - 0}{16} = \pi\,\text{cm}^2/\text{s}

    • AROC of Area with respect to Radius (t=0t = 0 to t=16 st = 16\,\text{s}):     r(0)=0=0 cmr(0) = \sqrt{0} = 0\,\text{cm}, r(16)=16=4 cmr(16) = \sqrt{16} = 4\,\text{cm}     AROC=A(16)−A(0)r(16)−r(0)=16π−04−0=16π4=4π cm2/cm\text{AROC} = \frac{A(16) - A(0)}{r(16) - r(0)} = \frac{16\pi - 0}{4 - 0} = \frac{16\pi}{4} = 4\pi\,\text{cm}^2/\text{cm}

    • AROC of Area with respect to Circumference (t=0t = 0 to t=16 st = 16\,\text{s}):     C(t)=2πr(t)=2πt cmC(t) = 2\pi r(t) = 2\pi \sqrt{t}\,\text{cm}     C(0)=0 cmC(0) = 0\,\text{cm}, C(16)=2π(4)=8π cmC(16) = 2\pi(4) = 8\pi\,\text{cm}     AROC=A(16)−A(0)C(16)−C(0)=16π−08π−0=2 cm2/cm\text{AROC} = \frac{A(16) - A(0)}{C(16) - C(0)} = \frac{16\pi - 0}{8\pi - 0} = 2\,\text{cm}^2/\text{cm}

  • Free-Falling Object Kinematics:

    • Platform height: 50 m50\,\text{m}. Initial velocity: 0 m/s0\,\text{m/s}.

    • Height equation: h(t)=−4.9t2+50 mh(t) = -4.9t^2 + 50\,\text{m}.

    • Average rates of change over shrinking intervals starting at t=2 st = 2\,\text{s}:

    • Interval [2,2.5][2, 2.5]:       AROC=h(2.5)−h(2)2.5−2=(−4.9(2.5)2+50)−(−4.9(2)2+50)0.5=−30.625+19.60.5=−11.0250.5=−22.05 m/s\text{AROC} = \frac{h(2.5) - h(2)}{2.5 - 2} = \frac{(-4.9(2.5)^2 + 50) - (-4.9(2)^2 + 50)}{0.5} = \frac{-30.625 + 19.6}{0.5} = \frac{-11.025}{0.5} = -22.05\,\text{m/s}

    • Interval [2,2.25][2, 2.25]:       AROC=h(2.25)−h(2)0.25=−24.80625+19.60.25=−20.825 m/s\text{AROC} = \frac{h(2.25) - h(2)}{0.25} = \frac{-24.80625 + 19.6}{0.25} = -20.825\,\text{m/s}

    • Interval [2,2.05][2, 2.05]:       AROC=h(2.05)−h(2)0.05=−19.845 m/s\text{AROC} = \frac{h(2.05) - h(2)}{0.05} = -19.845\,\text{m/s}

    • Interval [2,2.001][2, 2.001]:       AROC=h(2.001)−h(2)0.001=−19.6049 m/s\text{AROC} = \frac{h(2.001) - h(2)}{0.001} = -19.6049\,\text{m/s}

    • Suspected IROC at t=2 st = 2\,\text{s}: −19.6 m/s-19.6\,\text{m/s}.

  • General Projectile Motion:

    • Equation: h(t)=−12gt2+v0t+h0 mh(t) = -\frac{1}{2}gt^2 + v_0 t + h_0\,\text{m}, velocity v(t)=−gt+v0 m/sv(t) = -gt + v_0\,\text{m/s}.

    • AROC of height between t=0t = 0 and t=1t = 1:     AROC=h(1)−h(0)1−0=(−12g(1)2+v0(1)+h0)−h01=v0−12g m/s\text{AROC} = \frac{h(1) - h(0)}{1 - 0} = \frac{(-\frac{1}{2}g(1)^2 + v_0(1) + h_0) - h_0}{1} = v_0 - \frac{1}{2}g\,\text{m/s}

    • Time t=bt = b when velocity is zero:     v(t)=0  ⟹  −gt+v0=0  ⟹  t=b=v0g sv(t) = 0 \implies -gt + v_0 = 0 \implies t = b = \frac{v_0}{g}\,\text{s}

    • AROC of velocity between t=0t = 0 and t=bt = b:     AROC=v(b)−v(0)b−0=0−v0v0g−0=−v0v0g=−g m/s2\text{AROC} = \frac{v(b) - v(0)}{b - 0} = \frac{0 - v_0}{\frac{v_0}{g} - 0} = \frac{-v_0}{\frac{v_0}{g}} = -g\,\text{m/s}^2

  • Tree Growth Modeling:

    • Data table:

    • Diameter (cm\text{cm}): 1515, 4545, 7575, 120120

    • Height (m\text{m}): 1212, 2020, 2525, 3030

    • Secant line slopes from (15,12)(15, 12):

    • To (45,20)(45, 20): 20−1245−15=830=415≈0.267 m/cm\frac{20 - 12}{45 - 15} = \frac{8}{30} = \frac{4}{15} \approx 0.267\,\text{m/cm}

    • To (75,25)(75, 25): 25−1275−15=1360≈0.217 m/cm\frac{25 - 12}{75 - 15} = \frac{13}{60} \approx 0.217\,\text{m/cm}

    • To (120,30)(120, 30): 30−12120−15=18105=635≈0.171 m/cm\frac{30 - 12}{120 - 15} = \frac{18}{105} = \frac{6}{35} \approx 0.171\,\text{m/cm}

    • Secant line slopes from (45,20)(45, 20):

    • To (75,25)(75, 25): 25−2075−45=530=16≈0.167 m/cm\frac{25 - 20}{75 - 45} = \frac{5}{30} = \frac{1}{6} \approx 0.167\,\text{m/cm}

    • To (120,30)(120, 30): 30−20120−45=1075=215≈0.133 m/cm\frac{30 - 20}{120 - 45} = \frac{10}{75} = \frac{2}{15} \approx 0.133\,\text{m/cm}

    • Trend of IROC: Decreasing as diameter increases.

  • Economic Demand Function:

    • Demand model: D(s)=s2−600s+90000D(s) = s^2 - 600s + 90000 for 0≤s<3010 \le s < 301

    • Demand at price s=$0.50s = \$0.50 (50 cents):     D(0.5)=(0.5)2−600(0.5)+90000=0.25−300+90000=89699.25 booksD(0.5) = (0.5)^2 - 600(0.5) + 90000 = 0.25 - 300 + 90000 = 89699.25\,\text{books}

    • Price where demand is zero (D(s)=0D(s) = 0):     s2−600s+90000=0  ⟹  (s−300)2=0  ⟹  s=$300s^2 - 600s + 90000 = 0 \implies (s - 300)^2 = 0 \implies s = \$300

    • Instantaneous rate of change of demand at s=$10s = \$10:     Difference quotient:     D(s+h)−D(s)h=(s+h)2−600(s+h)+90000−(s2−600s+90000)h=2s+h−600\frac{D(s + h) - D(s)}{h} = \frac{(s + h)^2 - 600(s + h) + 90000 - (s^2 - 600s + 90000)}{h} = 2s + h - 600     Taking h→0h \to 0:     D′(s)=2s−600D'(s) = 2s - 600     At s=10s = 10:     D′(10)=2(10)−600=−580 books/$D'(10) = 2(10) - 600 = -580\,\text{books/\$}     Interpretation: At a price of $10\$10, increasing the book price by $1\$1 causes demand to decrease by approximately 580580 books.

  • Room Heater Temperature Model:

    • Model: T(m)=−110m2+4m+40T(m) = -\frac{1}{10}m^2 + 4m + 40 for 0≤m≤20 minutes0 \le m \le 20\,\text{minutes}.

    • Average rates of change:

    • Entire 20 minutes ([0,20][0, 20]):       AROC=T(20)−T(0)20−0=80−4020=2 ∘F/min\text{AROC} = \frac{T(20) - T(0)}{20 - 0} = \frac{80 - 40}{20} = 2\,^{\circ}\text{F/min}

    • First 10 minutes ([0,10][0, 10]):       AROC=T(10)−T(0)10−0=70−4010=3 ∘F/min\text{AROC} = \frac{T(10) - T(0)}{10 - 0} = \frac{70 - 40}{10} = 3\,^{\circ}\text{F/min}

    • Second 10 minutes ([10,20][10, 20]):       AROC=T(20)−T(10)20−10=80−7010=1 ∘F/min\text{AROC} = \frac{T(20) - T(10)}{20 - 10} = \frac{80 - 70}{10} = 1\,^{\circ}\text{F/min}

    • Instantaneous rates of change:

    • Difference quotient limit:       T′(m)=−15m+4T'(m) = -\frac{1}{5}m + 4

    • At m=5 minm = 5\,\text{min}: T′(5)=−15(5)+4=3 ∘F/minT'(5) = -\frac{1}{5}(5) + 4 = 3\,^{\circ}\text{F/min}

    • At m=10 minm = 10\,\text{min}: T′(10)=−15(10)+4=2 ∘F/minT'(10) = -\frac{1}{5}(10) + 4 = 2\,^{\circ}\text{F/min}

    • At m=15 minm = 15\,\text{min}: T′(15)=−15(15)+4=1 ∘F/minT'(15) = -\frac{1}{5}(15) + 4 = 1\,^{\circ}\text{F/min}

  • 2D Parametric Particle Kinematics:

    • Position vector: p(t)=(x(t),y(t))=(t,3t+5)p(t) = (x(t), y(t)) = (t, 3t + 5).

    • Instantaneous horizontal velocity:     vx(t)=lim⁡h→0(t+h)−th=lim⁡h→0hh=1v_x(t) = \lim_{h \to 0} \frac{(t + h) - t}{h} = \lim_{h \to 0} \frac{h}{h} = 1

    • Instantaneous vertical velocity:     vy(t)=lim⁡h→0(3(t+h)+5)−(3t+5)h=lim⁡h→03hh=3v_y(t) = \lim_{h \to 0} \frac{(3(t + h) + 5) - (3t + 5)}{h} = \lim_{h \to 0} \frac{3h}{h} = 3

  • Piecewise Driving Trip Model (Jenny's Trip):

    • Trip parameters:

    • Distance to coffee shop: 20 miles20\,\text{miles}.

    • Segment 1: Drives 25 mph25\,\text{mph} for 10 min10\,\text{min} (16 hour\frac{1}{6}\,\text{hour}). Distance covered = 25×16=256≈4.1667 miles25 \times \frac{1}{6} = \frac{25}{6} \approx 4.1667\,\text{miles}.

    • Segment 2: Idles for 5 min5\,\text{min} (112 hour\frac{1}{12}\,\text{hour}) on side of road. Distance from home remains 4.1667 miles4.1667\,\text{miles}.

    • Segment 3: Drives home at 35 mph35\,\text{mph}. Time taken = 4.166735=542 hours≈0.119 hours\frac{4.1667}{35} = \frac{5}{42}\,\text{hours} \approx 0.119\,\text{hours} (7.14 min7.14\,\text{min}).

    • Segment 4: Searches home for 5 min5\,\text{min} (112 hour\frac{1}{12}\,\text{hour}). Distance from home = 0 miles0\,\text{miles}.

    • Segment 5: Drives to coffee shop (20 miles20\,\text{miles}) at 40 mph40\,\text{mph}. Time taken = 2040=0.5 hours\frac{20}{40} = 0.5\,\text{hours} (30 min30\,\text{min}).

    • Calculations:

    • Total time tfinal=16+112+542+112+12=205210≈0.9733 hourst_{\text{final}} = \frac{1}{6} + \frac{1}{12} + \frac{5}{42} + \frac{1}{12} + \frac{1}{2} = \frac{205}{210} \approx 0.9733\,\text{hours} (58.4 minutes58.4\,\text{minutes}).

    • Average velocity over first 5 minutes: 25 mph25\,\text{mph}.

    • Average velocity over first 15 minutes:       Average Velocity=d(0.25)−d(0)0.25−0=4.16670.25=16.67 mph\text{Average Velocity} = \frac{d(0.25) - d(0)}{0.25 - 0} = \frac{4.1667}{0.25} = 16.67\,\text{mph}

    • Average velocity for entire trip:       Average Velocity=20−00.9733−0≈20.55 mph\text{Average Velocity} = \frac{20 - 0}{0.9733 - 0} \approx 20.55\,\text{mph}

    • Average speed for entire trip:       Total Distance=4.1667+4.1667+20=28.3333 miles\text{Total Distance} = 4.1667 + 4.1667 + 20 = 28.3333\,\text{miles}       Average Speed=28.33330.9733≈29.11 mph\text{Average Speed} = \frac{28.3333}{0.9733} \approx 29.11\,\text{mph}

Limits and One-Sided Limits

  • Definition of Limit:

    • The limit of f(x)f(x) as xx approaches cc, written lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L, exists if and only if both one-sided limits exist and are equal:     lim⁡x→c−f(x)=lim⁡x→c+f(x)=L\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L

  • Algebraic and Direct Substitution Limits:

    • Polynomial / Rational Direct Substitution:     lim⁡x→1x2+3x+2x+1=12+3(1)+21+1=62=3\lim_{x \to 1} \frac{x^2 + 3x + 2}{x + 1} = \frac{1^2 + 3(1) + 2}{1 + 1} = \frac{6}{2} = 3

    • Radical Direct Substitution:     lim⁡x→−1x2x+2=(−1)2−1+2=11=1\lim_{x \to -1} \frac{x^2}{\sqrt{x + 2}} = \frac{(-1)^2}{\sqrt{-1 + 2}} = \frac{1}{\sqrt{1}} = 1

    • Factoring Rational Functions:     lim⁡x→−3x2−9x+3=lim⁡x→−3(x−3)(x+3)x+3=lim⁡x→−3(x−3)=−3−3=−6\lim_{x \to -3} \frac{x^2 - 9}{x + 3} = \lim_{x \to -3} \frac{(x - 3)(x + 3)}{x + 3} = \lim_{x \to -3} (x - 3) = -3 - 3 = -6

    • Rationalizing / Complex Fraction Simplification:

    • Example 1:       lim⁡x→11x−1x−1=lim⁡x→11−xx(x−1)(x+1)=lim⁡x→1−1x(x+1)=−11(1+1)=−12\lim_{x \to 1} \frac{\frac{1}{\sqrt{x}} - 1}{x - 1} = \lim_{x \to 1} \frac{\frac{1 - \sqrt{x}}{\sqrt{x}}}{(\sqrt{x} - 1)(\sqrt{x} + 1)} = \lim_{x \to 1} \frac{-1}{\sqrt{x}(\sqrt{x} + 1)} = \frac{-1}{1(1 + 1)} = -\frac{1}{2}

    • Example 2:       lim⁡t→21t−12t−2=lim⁡t→22−t2tt−2=lim⁡t→2−(t−2)2t(t−2)=lim⁡t→2−12t=−14\lim_{t \to 2} \frac{\frac{1}{t} - \frac{1}{2}}{t - 2} = \lim_{t \to 2} \frac{\frac{2 - t}{2t}}{t - 2} = \lim_{t \to 2} \frac{-(t - 2)}{2t(t - 2)} = \lim_{t \to 2} \frac{-1}{2t} = -\frac{1}{4}

  • One-Sided Limits in Piecewise Functions:

    • For f(x)={3x+1if x≤02if x>0f(x) = \begin{cases} 3x + 1 & \text{if } x \le 0 \\ 2 & \text{if } x > 0 \end{cases}:

    • Left-hand limit: lim⁡x→0−f(x)=3(0)+1=1\lim_{x \to 0^-} f(x) = 3(0) + 1 = 1

    • Right-hand limit: lim⁡x→0+f(x)=2\lim_{x \to 0^+} f(x) = 2

    • Two-sided limit: Since lim⁡x→0−f(x)≠lim⁡x→0+f(x)\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x), lim⁡x→0f(x)\lim_{x \to 0} f(x) Does Not Exist (DNE).

  • Graphical Limit Analysis:

  

Graph for Limit Analysis

  • Behavior at x=0x = 0:

    • lim⁡x→0−f(x)=3\lim_{x \to 0^-} f(x) = 3

    • lim⁡x→0+f(x)=−1\lim_{x \to 0^+} f(x) = -1

    • lim⁡x→0f(x)=DNE\lim_{x \to 0} f(x) = \text{DNE} (jump discontinuity)

    • f(0)=3f(0) = 3

  • Behavior at x=2x = 2:

    • lim⁡x→2−f(x)=1\lim_{x \to 2^-} f(x) = 1

    • lim⁡x→2+f(x)=1\lim_{x \to 2^+} f(x) = 1

    • lim⁡x→2f(x)=1\lim_{x \to 2} f(x) = 1

    • f(2)=−2f(2) = -2 (removable point discontinuity)

  • Behavior at x=−3x = -3:

    • lim⁡x→−3−f(x)=∞\lim_{x \to -3^-} f(x) = \infty

    • lim⁡x→−3+f(x)=−∞\lim_{x \to -3^+} f(x) = -\infty

    • lim⁡x→−3f(x)=DNE\lim_{x \to -3} f(x) = \text{DNE} (vertical asymptote)

    • f(−3)=DNEf(-3) = \text{DNE}

  • Behavior at x=−6x = -6:

    • lim⁡x→−6−f(x)=4\lim_{x \to -6^-} f(x) = 4

    • lim⁡x→−6+f(x)=4\lim_{x \to -6^+} f(x) = 4

    • lim⁡x→−6f(x)=4\lim_{x \to -6} f(x) = 4

    • f(6)=1f(6) = 1

Continuity and Removable Discontinuities

  • Definition of Continuity:

    • A function f(x)f(x) is continuous at x=cx = c if all three conditions are satisfied:

    1. f(c)f(c) is defined (cc is in the domain of ff).

    2. lim⁡x→cf(x)\lim_{x \to c} f(x) exists.

    3. lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c).

  • Types of Discontinuities:

    • Removable Discontinuity (Hole): Occurs when lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L exists, but either f(c)f(c) is undefined or f(c)≠Lf(c) \neq L. Redefining f(c)=Lf(c) = L removes the discontinuity.

    • Jump Discontinuity: Occurs when both one-sided limits exist as finite numbers, but lim⁡x→c−f(x)≠lim⁡x→c+f(x)\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x).

    • Infinite Discontinuity (Vertical Asymptote): Occurs when one or both one-sided limits tend toward ±∞\pm\infty.

  • Detailed Rational Function Continuity Analysis:

    • Function: f(x)=x2−9x3+2x2−9x−18f(x) = \frac{x^2 - 9}{x^3 + 2x^2 - 9x - 18}

  

Rational Function Graph

  • Factor numerator and denominator:     f(x)=(x−3)(x+3)(x−3)(x+2)(x+3)=1x+2for x≠3,−3f(x) = \frac{(x - 3)(x + 3)}{(x - 3)(x + 2)(x + 3)} = \frac{1}{x + 2} \quad \text{for } x \neq 3, -3

  • Points of interest where elementary continuity fails:

    • At x=3x = 3:

      • Limit: lim⁡x→3f(x)=13+2=15\lim_{x \to 3} f(x) = \frac{1}{3 + 2} = \frac{1}{5}.

      • Value: f(3)f(3) is undefined.

      • Classification: Removable discontinuity (hole at (3,1/5)(3, 1/5)).

    • At x=−3x = -3:

      • Limit: \lim_{x \to -3} f(x) = \frac{1}{-3 + 2} = -1$.\n * Value: f(-3) is undefined.\n * Classification: Removable discontinuity (hole at (-3, -1)).\n * At x = -2:\n * One-sided limits: \lim_{x \to -2^-} f(x) = -\inftyandand\lim_{x \to -2^+} f(x) = \infty\n * Classification: Non-removable infinite discontinuity (vertical asymptote at x = -2).\n\n* **Detailed Piecewise Function Continuity Analysis:**\n * Function g(x)$:     g(x)={xif x<−100if x=−10−x−20if −10<x≤040x−2if x>0g(x) = \begin{cases} x & \text{if } x < -10 \\ 0 & \text{if } x = -10 \\ -x - 20 & \text{if } -10 < x \le 0 \\ \frac{40}{x - 2} & \text{if } x > 0 \end{cases}

  

Piecewise Graph

  • Analysis at key boundary points:

    • At x=−10x = -10:

      • Left-hand limit: lim⁡x→−10−g(x)=−10\lim_{x \to -10^-} g(x) = -10

      • Right-hand limit: lim⁡x→−10+g(x)=−(−10)−20=−10\lim_{x \to -10^+} g(x) = -(-10) - 20 = -10

      • Two-sided limit: lim⁡x→−10g(x)=−10\lim_{x \to -10} g(x) = -10

      • Function value: g(−10)=0g(-10) = 0

      • Conclusion: Discontinuous because g(−10)≠lim⁡x→−10g(x)g(-10) \neq \lim_{x \to -10} g(x). This is a removable discontinuity.

    • At x=0x = 0:

      • Left-hand limit: lim⁡x→0−g(x)=−0−20=−20\lim_{x \to 0^-} g(x) = -0 - 20 = -20

      • Right-hand limit: lim⁡x→0+g(x)=400−2=−20\lim_{x \to 0^+} g(x) = \frac{40}{0 - 2} = -20

      • Function value: g(0)=−0−20=−20g(0) = -0 - 20 = -20

      • Conclusion: Continuous at x=0x = 0 because lim⁡x→0g(x)=g(0)=−20\lim_{x \to 0} g(x) = g(0) = -20

    • At x=2x = 2:

      • Division by zero in 40x−2\frac{40}{x - 2}.

      • One-sided limits: lim⁡x→2−g(x)=−∞\lim_{x \to 2^-} g(x) = -\infty and lim⁡x→2+g(x)=∞\lim_{x \to 2^+} g(x) = \infty

      • Conclusion: Non-removable infinite discontinuity (vertical asymptote at x=2x = 2).

Intermediate Value Theorem and Wholesale Pricing Functions

  • Intermediate Value Theorem (IVT):

    • Statement: If f(x)f(x) is continuous on the closed interval [a,b][a, b], and NN is any number strictly between f(a)f(a) and f(b)f(b), then there exists at least one number c∈(a,b)c \in (a, b) such that f(c)=Nf(c) = N

    • Application to Equation Solving:

    • To prove x6+5x3=1x^6 + 5x^3 = 1 has a solution on the interval I=(0,1)I = (0, 1):

    • Define f(x)=x6+5x3−1f(x) = x^6 + 5x^3 - 1

    • f(x)f(x) is a polynomial and is therefore continuous everywhere, including on [0,1][0, 1].

    • Evaluate at interval boundaries:       f(0)=06+5(0)3−1=−1<0f(0) = 0^6 + 5(0)^3 - 1 = -1 < 0       f(1)=16+5(1)3−1=1+5−1=5>0f(1) = 1^6 + 5(1)^3 - 1 = 1 + 5 - 1 = 5 > 0

    • Since f(0)=−1<0f(0) = -1 < 0 and f(1)=5>0f(1) = 5 > 0, by the IVT, there exists a value c∈(0,1)c \in (0, 1) such that f(c)=0f(c) = 0, which implies c6+5c3=1c^6 + 5c^3 = 1

  • Wholesale Tiered Pricing and Total Cost Functions:

    • Supplier pricing structure:

    • Purchases of 500500 or fewer CDs: $0.80\$0.80 per CD.

    • Purchases between 501501 and 10001000 CDs: $0.72\$0.72 per CD.

    • Purchases over 10001000 CDs: $0.65\$0.65 per CD.

    • Price per CD equation P(x)$:\n    P(x) = \begin{cases} 0.80 & \text{if } 0 < x \le 500 \ 0.72 & \text{if } 500 < x \le 1000 \ 0.65 & \text{if } x > 1000 \end{cases}\n * Total Cost equation C(x) = x \cdot P(x)$:     C(x)={0.80xif 0<x≤5000.72xif 500<x≤10000.65xif x>1000C(x) = \begin{cases} 0.80x & \text{if } 0 < x \le 500 \\ 0.72x & \text{if } 500 < x \le 1000 \\ 0.65x & \text{if } x > 1000 \end{cases}

  

CD Total Cost Graph

  • One-sided cost limits as order size approaches 10001000 CDs:

    • Left-hand limit (x→1000−x \to 1000^-):       lim⁡x→1000−C(x)=0.72(1000)=$720\lim_{x \to 1000^-} C(x) = 0.72(1000) = \$720

    • Right-hand limit (x→1000+x \to 1000^+):       lim⁡x→1000+C(x)=0.65(1000)=$650\lim_{x \to 1000^+} C(x) = 0.65(1000) = \$650

  • Economic implication: Ordering 10011001 CDs ($650.65\$650.65) costs significantly less than ordering 10001000 CDs ($720\$720) due to the jump discontinuity in the tiered pricing structure.