Mat 271 Final Exam Review Notes

Part I Problem Set 1

1. Limits and Continuity from a Graph

Given a graph of a function g(t)g(t), find the following values or state if they are undefined or do not exist:

  • a) g(5)g(5)

  • b) lim⁡t→5g(t)\lim_{t \to 5} g(t)

  • c) g(3)g(3)

  • d) lim⁡t→3−g(t)\lim_{t \to 3^-} g(t)

  • e) lim⁡t→3+g(t)\lim_{t \to 3^+} g(t)

  • f) lim⁡t→3g(t)\lim_{t \to 3} g(t)

  • g) g(2)g(2)

  • h) lim⁡t→2g(t)\lim_{t \to 2} g(t)

  • i) g(−2)g(-2)

  • j) lim⁡t→−2−g(t)\lim_{t \to -2^-} g(t)

  • k) lim⁡t→−2+g(t)\lim_{t \to -2^+} g(t)

  • l) lim⁡t→−2g(t)\lim_{t \to -2} g(t)

2. Continuity of a Piecewise Function

Find the value of the constant aa that makes the function continuous on (−∞,∞)(-\infty, \infty).

f(x)={x+2aamp;if xlt;1 ax2+5amp;if x≥1f(x) = \begin{cases} x + 2a & \text{if } x < 1 \ ax^2 + 5 & \text{if } x \geq 1 \end{cases}

To ensure continuity, the two parts of the function must have the same value at x=1x = 1. Thus:

1+2a=a(1)2+51 + 2a = a(1)^2 + 5
1+2a=a+51 + 2a = a + 5
a=4a = 4

3. Limit Definition of the Derivative

Use the limit definition of the derivative to find f′(x)f'(x) for f(x)=2x2+xf(x) = 2x^2 + x.

The limit definition of the derivative is:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

For f(x)=2x2+xf(x) = 2x^2 + x,

f(x+h)=2(x+h)2+(x+h)=2(x2+2xh+h2)+x+h=2x2+4xh+2h2+x+hf(x+h) = 2(x+h)^2 + (x+h) = 2(x^2 + 2xh + h^2) + x + h = 2x^2 + 4xh + 2h^2 + x + h

f′(x)=lim⁡<em>h→0(2x2+4xh+2h2+x+h)−(2x2+x)hf'(x) = \lim<em>{h \to 0} \frac{(2x^2 + 4xh + 2h^2 + x + h) - (2x^2 + x)}{h} f′(x)=lim⁡</em>h→04xh+2h2+hh=lim⁡h→0(4x+2h+1)=4x+1f'(x) = \lim</em>{h \to 0} \frac{4xh + 2h^2 + h}{h} = \lim_{h \to 0} (4x + 2h + 1) = 4x + 1

Problems 4-5: Find y'

Do not simplify.

4. Derivative of Arctangent Function

y=arctan⁡(x3−1)y = \arctan(x^3 - 1)
Using the chain rule:
y′=11+(x3−1)2⋅(3x2)y' = \frac{1}{1 + (x^3 - 1)^2} \cdot (3x^2)
y′=3x21+(x3−1)2y' = \frac{3x^2}{1 + (x^3 - 1)^2}

5. Product and Chain Rule

y=x2tan⁡(1x)y = x^2 \tan(\frac{1}{x})
Using the product rule and chain rule:
y′=2xtan⁡(1x)+x2sec⁡2(1x)⋅(−1x2)y' = 2x \tan(\frac{1}{x}) + x^2 \sec^2(\frac{1}{x}) \cdot (-\frac{1}{x^2})
y′=2xtan⁡(1x)−sec⁡2(1x)y' = 2x \tan(\frac{1}{x}) - \sec^2(\frac{1}{x})

6. Implicit Differentiation

Find dydx\frac{dy}{dx} by implicit differentiation and then evaluate the derivative at the point (0,0)(0,0).

sin⁡(x2+y)=x2+2x+3y\sin(x^2 + y) = x^2 + 2x + 3y
Differentiating both sides with respect to xx:
cos⁡(x2+y)⋅(2x+dydx)=2x+2+3dydx\cos(x^2 + y) \cdot (2x + \frac{dy}{dx}) = 2x + 2 + 3\frac{dy}{dx}
At the point (0,0)(0,0),
cos⁡(0)⋅(0+dydx)=0+2+3dydx\cos(0) \cdot (0 + \frac{dy}{dx}) = 0 + 2 + 3\frac{dy}{dx}
1⋅dydx=2+3dydx1 \cdot \frac{dy}{dx} = 2 + 3\frac{dy}{dx}
−2=2dydx-2 = 2\frac{dy}{dx}
dydx=−1\frac{dy}{dx} = -1

7. Local and Absolute Extrema

Let f(x)=xexf(x) = xe^x. Find the local minimum value(s), local maximum value(s), absolute minimum value, and absolute maximum value.

First, find the derivative:
f′(x)=ex+xex=ex(1+x)f'(x) = e^x + xe^x = e^x(1+x)
Set f′(x)=0f'(x) = 0 to find critical points:
ex(1+x)=0e^x(1+x) = 0
Since exe^x is never zero, 1+x=01+x = 0 which means x=−1x = -1
Now, find the second derivative:
f′′(x)=ex(1+x)+ex=ex(2+x)f''(x) = e^x(1+x) + e^x = e^x(2+x)
Evaluate f''(-1) = e^{-1}(2-1) = e^{-1} > 0, so x=−1x = -1 is a local minimum.
The local minimum value is f(−1)=−1⋅e−1=−1ef(-1) = -1 \cdot e^{-1} = -\frac{1}{e}.
As x→−∞x \to -\infty, f(x)→0f(x) \to 0, and as x→∞x \to \infty, f(x)→∞f(x) \to \infty. Thus, there is no absolute maximum.
The absolute minimum is −1e-\frac{1}{e} at x=−1x = -1.

8. Limit at Infinity

lim⁡<em>x→∞x+29x2+1\lim<em>{x \to \infty} \frac{x+2}{9x^2+1} Divide by the highest power of xx in the denominator, which is x2x^2: lim⁡</em>x→∞xx2+2x29x2x2+1x2=lim⁡<em>x→∞1x+2x29+1x2\lim</em>{x \to \infty} \frac{\frac{x}{x^2} + \frac{2}{x^2}}{\frac{9x^2}{x^2} + \frac{1}{x^2}} = \lim<em>{x \to \infty} \frac{\frac{1}{x} + \frac{2}{x^2}}{9 + \frac{1}{x^2}} As x→∞x \to \infty, 1x→0\frac{1}{x} \to 0 and 1x2→0\frac{1}{x^2} \to 0. lim⁡</em>x→∞0+09+0=09=0\lim</em>{x \to \infty} \frac{0 + 0}{9 + 0} = \frac{0}{9} = 0

9. Limit with a Radical

lim⁡<em>x→03+x−3x\lim<em>{x \to 0} \frac{\sqrt{3+x} - \sqrt{3}}{x} Multiply by the conjugate: lim⁡</em>x→03+x−3x⋅3+x+33+x+3=lim⁡<em>x→0(3+x)−3x(3+x+3)=lim⁡</em>x→0xx(3+x+3)\lim</em>{x \to 0} \frac{\sqrt{3+x} - \sqrt{3}}{x} \cdot \frac{\sqrt{3+x} + \sqrt{3}}{\sqrt{3+x} + \sqrt{3}} = \lim<em>{x \to 0} \frac{(3+x) - 3}{x(\sqrt{3+x} + \sqrt{3})} = \lim</em>{x \to 0} \frac{x}{x(\sqrt{3+x} + \sqrt{3})}
lim⁡x→013+x+3=13+0+3=123\lim_{x \to 0} \frac{1}{\sqrt{3+x} + \sqrt{3}} = \frac{1}{\sqrt{3+0} + \sqrt{3}} = \frac{1}{2\sqrt{3}}

10. Limit by Factoring

lim⁡<em>x→4x2−5x+4x2−2x−8\lim<em>{x \to 4} \frac{x^2 - 5x + 4}{x^2 - 2x - 8} Factor the numerator and denominator: lim⁡</em>x→4(x−4)(x−1)(x−4)(x+2)=lim⁡x→4x−1x+2=4−14+2=36=12\lim</em>{x \to 4} \frac{(x-4)(x-1)}{(x-4)(x+2)} = \lim_{x \to 4} \frac{x-1}{x+2} = \frac{4-1}{4+2} = \frac{3}{6} = \frac{1}{2}

Part I Problem Set 2

1. Limits and Function Values from a Graph

Given a graph of a function f(x)f(x), find the following values or state if they are undefined or do not exist.

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

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

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

  • d) f(−2)f(-2)

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

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

  • g) lim⁡x→1f(1)\lim_{x \to 1} f(1)

  • h) f(1)f(1)

  • i) f(−4)f(-4)

2. Limit of Absolute Value Function

Find lim⁡<em>x→0∣x∣x\lim<em>{x \to 0} \frac{|x|}{x}. lim⁡</em>x→0−∣x∣x=lim⁡<em>x→0−−xx=−1\lim</em>{x \to 0^-} \frac{|x|}{x} = \lim<em>{x \to 0^-} \frac{-x}{x} = -1 lim⁡</em>x→0+∣x∣x=lim⁡x→0+xx=1\lim</em>{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1
Since the left-hand limit and right-hand limit are not equal, the limit does not exist.

3. Derivative of a Polynomial and Exponential Function

Find f′(x)f'(x) for f(x)=x7−2x3+e2f(x) = x^7 - 2x^3 + e^2. Do not simplify.

f′(x)=7x6−6x2+0f'(x) = 7x^6 - 6x^2 + 0
f′(x)=7x6−6x2f'(x) = 7x^6 - 6x^2

4. Derivative of a Complex Function

Find f′(x)f'(x) for f(x)=x+eπcos⁡4+sin⁡5(6x)f(x) = \frac{x + e^{\pi}}{\cos^4 + \sin^5(6x)}. Do not simplify.

Using the quotient rule:

f′(x)=(cos⁡4(x)+sin⁡5(6x))(1)−(x+eπ)(4cos⁡3(x)(−sin⁡(x))+5sin⁡4(6x)(cos⁡(6x))(6))(cos⁡4(x)+sin⁡5(6x))2f'(x) = \frac{(\cos^4 (x) + \sin^5(6x))(1) - (x + e^{\pi})(4\cos^3(x)(-\sin(x)) + 5\sin^4(6x)(\cos(6x))(6))}{(\cos^4 (x) + \sin^5(6x))^2}

5. Logarithmic Differentiation

Use logarithmic differentiation to find f′(x)f'(x) for f(x)=(sin⁡(x))xf(x) = (\sin(x))^x. Do not simplify.

Take the natural logarithm of both sides:
ln⁡(f(x))=ln⁡((sin⁡(x))x)=xln⁡(sin⁡(x))\ln(f(x)) = \ln((\sin(x))^x) = x \ln(\sin(x))
Differentiate both sides with respect to xx:
f′(x)f(x)=ln⁡(sin⁡(x))+x⋅cos⁡(x)sin⁡(x)=ln⁡(sin⁡(x))+xcot⁡(x)\frac{f'(x)}{f(x)} = \ln(\sin(x)) + x \cdot \frac{\cos(x)}{\sin(x)} = \ln(\sin(x)) + x \cot(x)
Multiply both sides by f(x)f(x)
f′(x)=(sin⁡(x))x(ln⁡(sin⁡(x))+xcot⁡(x))f'(x) = (\sin(x))^x (\ln(\sin(x)) + x \cot(x))

6. Related Rates

A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/s4 \text{ ft/s}. After 1212 seconds, how rapidly is the area enclosed by the ripple increasing?

The area of a circle is A=πr2A = \pi r^2.
Differentiating with respect to time tt:
dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}
Given drdt=4 ft/s\frac{dr}{dt} = 4 \text{ ft/s}. After 1212 seconds, r=4⋅12=48 ftr = 4 \cdot 12 = 48 \text{ ft}.
dAdt=2π(48)(4)=384π ft2/s\frac{dA}{dt} = 2\pi (48)(4) = 384\pi \text{ ft}^2/\text{s}

7. Applications of Calculus

Given f(x)=3x4−4x3+2012f(x) = 3x^4 - 4x^3 + 2012. Use calculus to answer parts a – d.

a) Find the intervals where ff is increasing and decreasing.

First, find the derivative:

f′(x)=12x3−12x2=12x2(x−1)f'(x) = 12x^3 - 12x^2 = 12x^2(x-1)
Set f′(x)=0f'(x) = 0:
12x2(x−1)=012x^2(x-1) = 0
x=0,1x = 0, 1
Test intervals: (−∞,0)(-\infty, 0), (0,1)(0, 1), (1,∞)(1, \infty)
For (−∞,0)(-\infty, 0), let x=−1x = -1: f'(-1) = 12(-1)^2(-1-1) = -24 < 0 (decreasing) For (0,1)(0, 1), let x=0.5x = 0.5: f'(0.5) = 12(0.5)^2(0.5-1) = -1.5 < 0 (decreasing) For (1,∞)(1, \infty), let x=2x = 2: f'(2) = 12(2)^2(2-1) = 48 > 0 (increasing)

Decreasing: (−∞,1)(-\infty, 1)
Increasing: (1,∞)(1, \infty)

b) Find all local and absolute minimum and maximum values and the values of xx at which they occur.
Local minimum at x=1x = 1: f(1)=3(1)4−4(1)3+2012=3−4+2012=2011f(1) = 3(1)^4 - 4(1)^3 + 2012 = 3 - 4 + 2012 = 2011
Since f(x)→∞f(x) \to \infty as x→±∞x \to \pm \infty, there is no local or absolute maximum.
Absolute minimum at x=1x = 1: f(1)=2011f(1) = 2011

c) Find the intervals where ff is concave up and concave down.

Find the second derivative:

f′′(x)=36x2−24x=12x(3x−2)f''(x) = 36x^2 - 24x = 12x(3x - 2)
Set f′′(x)=0f''(x) = 0:
12x(3x−2)=012x(3x - 2) = 0
x=0,23x = 0, \frac{2}{3}
Test intervals: (−∞,0)(-\infty, 0), (0,23)(0, \frac{2}{3}), (23,∞)(\frac{2}{3}, \infty)
For (−∞,0)(-\infty, 0), let x=−1x = -1: f''(-1) = 12(-1)(3(-1) - 2) = 60 > 0 (concave up)
For (0,23)(0, \frac{2}{3}), let x=12x = \frac{1}{2}: f''(\frac{1}{2}) = 12(\frac{1}{2})(3(\frac{1}{2}) - 2) = -3 < 0 (concave down) For (23,∞)(\frac{2}{3}, \infty), let x=1x = 1: f''(1) = 12(1)(3(1) - 2) = 12 > 0 (concave up)

Concave Up: (−∞,0)∪(23,∞)(-\infty, 0) \cup (\frac{2}{3}, \infty)
Concave Down: (0,23)(0, \frac{2}{3})

d) Find all points of inflection. (Round to thousandths)
Points of inflection occur where the concavity changes, which is at x=0x = 0 and x=23x = \frac{2}{3}.
At x=0x = 0, f(0)=3(0)4−4(0)3+2012=2012f(0) = 3(0)^4 - 4(0)^3 + 2012 = 2012
At x=23x = \frac{2}{3}, f(23)=3(23)4−4(23)3+2012=3(1681)−4(827)+2012=1627−3227+2012=−1627+2012≈2011.407f(\frac{2}{3}) = 3(\frac{2}{3})^4 - 4(\frac{2}{3})^3 + 2012 = 3(\frac{16}{81}) - 4(\frac{8}{27}) + 2012 = \frac{16}{27} - \frac{32}{27} + 2012 = -\frac{16}{27} + 2012 \approx 2011.407

Points of inflection: (0,2012)(0, 2012) and (23,2011.407)(\frac{2}{3}, 2011.407)