Calculus BC Exam Review: Limits and Derivatives

Vocabulary practice flashcards covering limits, continuity, derivative rules, tangent lines, the Intermediate Value Theorem, and the Squeeze Theorem from the Calculus BC exam review.

Limit of $\lim_{x \to 5} \frac{x^3 - 9x^2 + 20x}{(x-5)(x-2)}$

$\frac{5}{3}$, found by factoring the numerator to $x(x-5)(x-4)$, canceling $(x-5)$, and substituting $x = 5$ into $\frac{x(x-4)}{x-2}$.

Limit of $\lim_{x \to -\infty} (x^6 - 2x^4 + 3x - 1)$

$\infty$, because as $x \to -\infty$, the highest degree term $x^6$ dominates and $(-\infty)^6 \to \infty$.

One-Sided Limit of $\lim_{x \to -7^+} \frac{(x+7)(x-9)}{(x+7)^2}$

$-\infty$, obtained by simplifying the expression to $\frac{x-9}{x+7}$, where the numerator approaches $-16$ and the denominator approaches $0^+$ as $x \to -7^+$.

Limit of $\lim_{x \to \infty} \frac{-x^2 + 3x^8 + 2x + 1}{x^3 - 6x^8 + x + 2}$

$-\frac{1}{2}$, determined by evaluating the ratio of the coefficients of the highest power term ($x^8$), which gives $\frac{3}{-6}$.

Intervals with roots for continuous function $f(x)$ in the table

The intervals $(-3, 0)$, $(0, 1)$, and $(7, 10)$, because $f(x)$ changes signs across each of these intervals, satisfying the Intermediate Value Theorem.

Limit Definition of Derivative for $f(x) = -3x^2 + x + 1$

$f'(x) = \lim_{h \to 0} \frac{-3(x+h)^2 + (x+h) + 1 - (-3x^2 + x + 1)}{h} = -6x + 1$.

Derivative of $f(x) = 3x^7 - x + \frac{4}{x^7}$

$f'(x) = 21x^6 - 1 - \frac{28}{x^8}$, calculated by rewriting $\frac{4}{x^7}$ as $4x^{-7}$ and applying the power rule.

Derivative of $g(x) = \tan(x) - \cos(x) \ln(x)$

$g'(x) = \sec^2(x) + \sin(x) \ln(x) - \frac{\cos(x)}{x}$, derived using basic derivative rules and the product rule on $\cos(x) \ln(x)$.

Derivative of $h(x) = (x^5 - x^4) e^x$

$h'(x) = (x^5 + 4x^4 - 4x^3) e^x$, evaluated using the product rule: $(5x^4 - 4x^3) e^x + (x^5 - x^4) e^x$.

Derivative of $j(x) = 3 - x + 3 + \sin(x^2 + x)$

$j'(x) = -1 + (2x + 1) \cos(x^2 + x)$, computed using sum rules and applying the chain rule to $\sin(x^2 + x)$.

Tangent Line of $f(x) = x^3 - x + 1$ at $x = 1$

$y = 2x - 1$, derived using the point $(1, 1)$ and slope $f'(1) = 3(1)^2 - 1 = 2$ in point-slope form $y - 1 = 2(x - 1)$.

Intermediate Value Theorem (IVT)

A theorem stating that if a function $f$ is continuous on a closed interval $[a, b]$, it takes on every value between $f(a)$ and $f(b)$ on that interval.

IVT Applicability to $x^2 - 3x + \sin(x) = 4$ on $(-2, 2)$

Justifiable, because $k(x) = x^2 - 3x + \sin(x)$ is continuous on $[-2, 2]$, with $k(-2) = 10 - \sin(2) > 4$ and $k(2) = -2 + \sin(2) < 4$.

IVT Applicability to $f(x) = \frac{x+1}{x} - 1$ on $(-1, 1)$

Not justifiable using IVT, because $f(x) = \frac{1}{x}$ is discontinuous at $x = 0$, failing the continuity condition on $(-1, 1)$.

Squeeze Theorem

A theorem stating that if $g(x) \le f(x) \le h(x)$ near $c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$.

Squeeze Theorem Application to $-x^4 \le g(x) \le x^4$ as $x \to 0$

Valid, because both $\lim_{x \to 0} (-x^4) = 0$ and $\lim_{x \to 0} (x^4) = 0$, which squeezes $\lim_{x \to 0} g(x)$ to equal $0$.