AP Calculus BC Quiz: Local Linearity And Linearization

20 questions · exam conditions

0:00

Local Linearity And LinearizationQuestion 1 of 20

A temperature model $T$ satisfies $T(10)=68$ and $T'(10)=1.5$ . What is the linear approximation near $t=10$ ?

L(t)=68+1.5(t-10)

L(t)=10+1.5(t-68)

L(t)=68-1.5(t-10)

L(t)=68+10(t-1.5)

L(t)=1.5+68(t-10)

← Back to quizzes

AP Calculus BC Quiz

AP Calculus BC Quiz: Local Linearity And Linearization

Practice Local Linearity And Linearization in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Local Linearity And Linearization, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A temperature model $T$ satisfies $T(10)=68$ and $T'(10)=1.5$ . What is the linear approximation near $t=10$ ?

$L(t)=68+1.5(t-10)$ (correct answer)
$L(t)=10+1.5(t-68)$
$L(t)=68-1.5(t-10)$
$L(t)=68+10(t-1.5)$
$L(t)=1.5+68(t-10)$

Explanation: This question involves linearization of a temperature model at a specific time. The linear approximation formula L(t) = T(a) + T'(a)(t - a) applies here with a = 10. Given T(10) = 68 and T'(10) = 1.5, we substitute these values directly. The linearization becomes L(t) = 68 + 1.5(t - 10), representing how temperature changes near t = 10. This tangent line approximation uses the temperature value 68 and rate of change 1.5 at time t = 10. Choice B swaps the function value and point of linearization, which would be incorrect. Remember: linearization format is always function_value + derivative × (input - point).

Question 2

For $f$ , $f(0)=1$ and $f'(0)=-frac12$ . What is the linear approximation near $x=0$ ?

$L(x)=1-\tfrac12(x-0)$ (correct answer)
$L(x)=0-\tfrac12(x-1)$
$L(x)=1+\tfrac12(x-0)$
$L(x)=-\tfrac12+1(x-0)$
$L(x)=1-0(x+\tfrac12)$

Explanation: This linearization problem involves fractional derivative and centers at x = 0. Using L(x) = f(a) + f'(a)(x - a) with a = 0, f(0) = 1, and f'(0) = -1/2. The linearization at the origin simplifies to L(x) = 1 + (-1/2)(x - 0) = 1 - (1/2)x. Since we're linearizing at x = 0, the (x - 0) term becomes just x. This tangent line starts at (0, 1) and has slope -1/2, decreasing as x increases. Choice C uses +1/2 instead of -1/2, giving an incorrect positive slope. When linearizing at x = 0, the formula reduces to f(0) + f'(0)x.

Question 3

A function $M$ has $M(\pi)=2\pi$ and $M'(\pi)=2$ . What is the linear approximation near $x=\pi$ ?

$L(x)=2\pi+2(x-\pi)$ (correct answer)
$L(x)=\pi+2(x-2\pi)$
$L(x)=2\pi-2(x-\pi)$
$L(x)=2+2\pi(x-\pi)$
$L(x)=2\pi+\pi(x-2)$

Explanation: This linearization involves

\pi

values with positive integer derivative. Using

L(x) = M(a) + M'(a)(x - a)

with a =

\pi

, M(

\pi

) = 2

\pi

, and M'(

\pi

) = 2. The linear approximation becomes

L(x) = 2\pi + 2(x - \pi)

. The tangent line passes through (

\pi

, 2

\pi

) with slope 2, increasing by 2 units per unit increase in x. The function value is twice the linearization point, creating interesting geometric relationships. Choice C uses -2 instead of +2, representing decreasing behavior when the derivative indicates increase. When working with

\pi

values, maintain exact forms and verify derivative signs for proper function behavior.

Question 4

For differentiable $F$ , $F(9)=1$ and $F'(9)=8$ . Which is the tangent line at $x=9$ ?

$y=1+8(x-9)$ (correct answer)
$y=9+8(x-1)$
$y=1-8(x-9)$
$y=8+1(x-9)$
$y=1+9(x-8)$

Explanation: This question asks for the tangent line of function F with specific numerical values. Using y = F(a) + F'(a)(x - a) with a = 9, F(9) = 1, and F'(9) = 8. The tangent line equation becomes y = 1 + 8(x - 9). This line passes through (9, 1) with slope 8, indicating rapid increase of 8 units per unit increase in x. The linearization provides an excellent approximation of F(x) near x = 9 due to the substantial slope. Choice B incorrectly uses (x - 1) instead of (x - 9), representing linearization at the wrong point. Verify the linearization point matches the given x-value where function value and derivative are specified.

Question 5

For differentiable $f$ , given $f(2)=5$ and $f'(2)=-3$ , what is the linearization of $f$ at $x=2$ ?

$L(x)=5-3(x-2)$ (correct answer)
$L(x)=5-3(x+2)$
$L(x)=2-3(x-5)$
$L(x)=5+3(x-2)$
$L(x)=5-2(x-3)$

Explanation: This question requires finding the linearization of a function at a specific point. The linearization formula is L(x) = f(a) + f'(a)(x - a), where a is the point of linearization. Given f(2) = 5 and f'(2) = -3, we substitute a = 2 into the formula. This gives us L(x) = 5 + (-3)(x - 2) = 5 - 3(x - 2). The tangent line approximates the function near x = 2 using the function value and slope at that point. Choice B is tempting because it uses the correct values but has (x + 2) instead of (x - 2), which would represent linearization at x = -2. Always verify the linearization formula: start with f(a), add f'(a) times (x - a).

Question 6

A function $S$ has $S(11)=\!-2$ and $S'(11)=9$ . Which is the tangent line at $x=11$ ?

$y=-2+9(x-11)$ (correct answer)
$y=11+9(x+2)$
$y=-2-9(x-11)$
$y=9-2(x-11)$
$y=-2+11(x-9)$

Explanation: This linearization involves large numerical values with positive derivative. Using y = S(a) + S'(a)(x - a) with a = 11, S(11) = -2, and S'(11) = 9. The tangent line becomes y = -2 + 9(x - 11). Despite the negative function value, the large positive derivative indicates rapid increase. The tangent line passes through (11, -2) and increases by 9 units per unit increase in x, quickly overcoming the negative starting value. Choice C uses -9 instead of +9, representing continued decrease rather than the strong upward trend. Large positive derivatives can overcome negative function values to create increasing linearizations.

Question 7

Let $k(7)=\tfrac{9}{2}$ and $k'(7)=-\tfrac{1}{3}$ . Which is the tangent line at $x=7$ ?

$y=\tfrac{9}{2}-\tfrac{1}{3}(x-7)$ (correct answer)
$y=7-\tfrac{1}{3}(x-\tfrac{9}{2})$
$y=\tfrac{9}{2}+\tfrac{1}{3}(x-7)$
$y=-\tfrac{1}{3}+\tfrac{9}{2}(x-7)$
$y=\tfrac{9}{2}-7(x+\tfrac{1}{3})$

Explanation: This linearization involves fractional values with attention to negative derivatives. Using y = k(a) + k'(a)(x - a) with a = 7, k(7) = 9/2, and k'(7) = -1/3. The tangent line becomes y = 9/2 + (-1/3)(x - 7) = 9/2 - (1/3)(x - 7). The line passes through (7, 9/2) with slope -1/3, decreasing slowly as x increases. The negative derivative indicates the function decreases by 1 unit for every 3 units increase in x. Choice C uses +1/3 instead of -1/3, representing increasing behavior when the function is actually decreasing. Carefully track signs when working with fractional derivatives to ensure correct increasing/decreasing behavior.

Question 8

Given $p(4)=7$ and $p'(4)=2$ , which is the tangent line to $p$ at $x=4$ ?

$y=7+2(x-4)$ (correct answer)
$y=4+2(x-7)$
$y=7-2(x-4)$
$y=2+7(x-4)$
$y=7+4(x-2)$

Explanation: This question seeks the tangent line equation using the standard linearization approach. With p(4) = 7 and p'(4) = 2, we apply the formula y = f(a) + f'(a)(x - a) at a = 4. Substituting the given values yields y = 7 + 2(x - 4). This tangent line passes through the point (4, 7) with slope 2, providing the best linear approximation of p near x = 4. The linear function captures both the function value and instantaneous rate of change at the point of tangency. Choice B incorrectly reverses the function value and point, using y = 4 + 2(x - 7) instead. Standard linearization checklist: correct point a, function value f(a), derivative f'(a), and format f(a) + f'(a)(x - a).

Question 9

Let $r(1)=\tfrac12$ and $r'(1)=\tfrac12$ . What is the linear approximation of $r$ at $x=1$ ?

$L(x)=\tfrac12+\tfrac12(x-1)$ (correct answer)
$L(x)=1+\tfrac12(x-\tfrac12)$
$L(x)=\tfrac12-\tfrac12(x-1)$
$L(x)=\tfrac12+1(x-\tfrac12)$
$L(x)=\tfrac12+\tfrac12(x+1)$

Explanation: This linearization problem involves fractional values for both function value and derivative. Using L(x) = r(a) + r'(a)(x - a) with a = 1, r(1) = 1/2, and r'(1) = 1/2. The linear approximation becomes L(x) = 1/2 + (1/2)(x - 1). Both the function value and slope are 1/2, creating a tangent line that passes through (1, 1/2) with slope 1/2. This means the function increases by 1 unit for every 2 units of increase in x. Choice C uses -1/2 for the derivative, which would represent decreasing behavior. When both function value and derivative are fractions, maintain precision and verify signs throughout the calculation.

Question 10

A concentration function $C$ satisfies $C(0)=8$ and $C'(0)=5$ . What is the linear approximation near $t=0$ ?

$L(t)=8+5t$ (correct answer)
$L(t)=0+5(t-8)$
$L(t)=8-5t$
$L(t)=5+8t$
$L(t)=8+0(t-5)$

Explanation: This problem asks for linearization at x = 0, which simplifies the algebraic form. Using L(t) = C(a) + C'(a)(t - a) with a = 0, C(0) = 8, and C'(0) = 5. The linearization becomes L(t) = 8 + 5(t - 0) = 8 + 5t. When linearizing at the origin, the (t - 0) term simplifies to just t, creating a clean linear function. The concentration starts at 8 when t = 0 and increases by 5 units per unit increase in time. Choice B incorrectly maintains the (t - 8) form, suggesting linearization at t = 8 rather than t = 0. Linearization at the origin produces the simple form f(0) + f'(0)x.

Question 11

A function $a$ has $a(8)=-6$ and $a'(8)=0.2$ . What is the linearization at $x=8$ ?

$L(x)=-6+0.2(x-8)$ (correct answer)
$L(x)=8+0.2(x+6)$
$L(x)=-6-0.2(x-8)$
$L(x)=0.2-6(x-8)$
$L(x)=-6+8(x-0.2)$

Explanation: This linearization problem features a decimal derivative and negative function value. Using L(x) = a(b) + a'(b)(x - b) with b = 8, a(8) = -6, and a'(8) = 0.2. The linearization becomes L(x) = -6 + 0.2(x - 8), representing the tangent line at (8, -6). The positive derivative 0.2 indicates the function is increasing slowly near x = 8, rising by 0.2 units per unit increase in x. Even though the function value is negative, the positive derivative shows upward movement. Choice C uses -0.2 instead of +0.2, suggesting decreasing behavior rather than increasing. Check that derivative signs match the intended function behavior (increasing vs decreasing).

Question 12

Given $b(\pi)=0$ and $b'(\pi)=-2$ , which is the linear approximation near $x=\pi$ ?

$L(x)=-2(x-\pi)$ (correct answer)
$L(x)=-2(x+\pi)$
$L(x)=2(x-\pi)$
$L(x)=\pi-2(x-0)$
$L(x)=0-\pi(x+2)$

Explanation: This problem involves linearization at x =

\pi

with a zero function value. Using

L(x) = b(a) + b'(a)(x - a)

with a =

\pi

, b(

\pi

) = 0, and b'(

\pi

) = -2. The linearization becomes

L(x) = 0 + (-2)(x - \pi) = -2(x - \pi)

. Since the function value at

\pi

is zero, the tangent line passes through (

\pi

, 0) with slope -2. This represents a decreasing linear function that crosses the x-axis at x =

\pi

. Choice C uses +2 instead of -2, which would represent increasing behavior rather than decreasing. When the function value is zero, the linearization simplifies to just the derivative times

(x - a)

Question 13

If $m(5)=-1$ and $m'(5)=3$ , what is the tangent line equation at $x=5$ ?

$y=-1+3(x-5)$ (correct answer)
$y=5+3(x+1)$
$y=-1-3(x-5)$
$y=3-1(x-5)$
$y=-1+5(x-3)$

Explanation: This linearization question follows the standard tangent line format. With m(5) = -1 and m'(5) = 3, we use the formula y = f(a) + f'(a)(x - a) at a = 5. Substituting gives y = -1 + 3(x - 5), which represents the tangent line at the point (5, -1). The linear approximation has slope 3, meaning the function increases by 3 units for each unit increase in x near x = 5. This tangent line provides the best linear estimate of m(x) in a neighborhood around x = 5. Choice C uses -3 instead of +3 for the slope, which would represent a decreasing function rather than increasing. Always verify: point coordinates, slope sign, and linearization point in the (x - a) term.

Question 14

If $g(1)=-2$ and $g'(1)=4$ , which equation is the tangent line to $g$ at $x=1$ ?

$y=-2+4(x-1)$ (correct answer)
$y=1+4(x+2)$
$y=-2-4(x-1)$
$y=4-2(x-1)$
$y=-2+1(x-4)$

Explanation: This problem asks for the tangent line equation, which is the same as linearization. The tangent line at x = a has equation y = f(a) + f'(a)(x - a). With g(1) = -2 and g'(1) = 4, we substitute a = 1 to get the equation. This yields y = -2 + 4(x - 1), which represents the linear approximation of g near x = 1. The tangent line passes through (1, -2) with slope 4, making it the best linear approximation in the neighborhood of x = 1. Choice B incorrectly uses (x + 2) instead of (x - 1), suggesting linearization at a different point. To check linearization: identify the point a, use f(a) as the constant term, and f'(a)(x - a) as the linear term.

Question 15

A function $J$ satisfies $J(\tfrac12)=1$ and $J'(\tfrac12)=6$ . What is the linear approximation near $x=\tfrac12$ ?

$L(x)=1+6\left(x-\tfrac12\right)$ (correct answer)
$L(x)=\tfrac12+6(x-1)$
$L(x)=1-6\left(x-\tfrac12\right)$
$L(x)=6+1\left(x-\tfrac12\right)$
$L(x)=1+\tfrac12(x-6)$

Explanation: This linearization involves a fractional x-value with positive integer derivative. Using L(x) = J(a) + J'(a)(x - a) with a = 1/2, J(1/2) = 1, and J'(1/2) = 6. The linear approximation becomes L(x) = 1 + 6(x - 1/2). The tangent line passes through (1/2, 1) with slope 6, increasing rapidly by 6 units per unit increase in x. Linearizing at fractional points requires careful handling of the (x - a) term. Choice C uses -6 instead of +6, representing decreasing behavior when the positive derivative indicates increase. When linearizing at fractional points, maintain the exact fractional form in (x - a).

Question 16

If $c(-3)=10$ and $c'(-3)=7$ , what is the tangent line to $c$ at $x=-3$ ?

$y=10+7(x+3)$ (correct answer)
$y=-3+7(x-10)$
$y=10-7(x+3)$
$y=7+10(x+3)$
$y=10+3(x-7)$

Explanation: This linearization takes place at a negative x-value with careful attention to the algebraic form. Using y = c(a) + c'(a)(x - a) with a = -3, c(-3) = 10, and c'(-3) = 7. The tangent line becomes y = 10 + 7(x - (-3)) = 10 + 7(x + 3). The expression (x - (-3)) simplifies to (x + 3), which is the standard form when linearizing at negative points. This tangent line passes through (-3, 10) with slope 7, showing rapid increase. Choice C uses -7 instead of +7, representing decreasing rather than increasing behavior. Remember that (x - a) becomes (x + |a|) when a is negative.

Question 17

Let $h$ be differentiable with $h(-1)=3$ and $h'(-1)=0$ . What is the linearization at $x=-1$ ?

$L(x)=3$ (correct answer)
$L(x)=3+(x+1)$
$L(x)=0+3(x+1)$
$L(x)=3-(x+1)$
$L(x)=3+0(x-1)$

Explanation: This linearization problem has a special case where the derivative equals zero. Using the formula L(x) = h(a) + h'(a)(x - a) with a = -1, h(-1) = 3, and h'(-1) = 0. When the derivative is zero, the linearization simplifies significantly. Substituting gives L(x) = 3 + 0(x - (-1)) = 3 + 0(x + 1) = 3. The tangent line is horizontal at y = 3, meaning the function is approximately constant near x = -1. Choice E shows L(x) = 3 + 0(x - 1), which uses the wrong point for linearization. When the derivative is zero at a point, the linearization is simply the constant function value.

Question 18

A position function $s$ has $s(3)=20$ and $s'(3)=6$ . Which linearization approximates $s(x)$ near $x=3$ ?

$L(x)=20+6(x-3)$ (correct answer)
$L(x)=3+6(x-20)$
$L(x)=20-6(x-3)$
$L(x)=6+20(x-3)$
$L(x)=20+3(x-6)$

Explanation: This problem asks for linearization of a position function, applying physics context to calculus. Using the standard formula L(x) = s(a) + s'(a)(x - a) with a = 3, s(3) = 20, and s'(3) = 6. The linearization becomes L(x) = 20 + 6(x - 3), representing position as a linear function near t = 3. The tangent line shows the position starts at 20 when x = 3, with velocity (derivative) of 6 units per time unit. This linear approximation captures both the instantaneous position and velocity at the specific time. Choice C uses -6 instead of +6, which would represent decreasing position rather than increasing. Linearization checklist: position value, velocity (derivative), correct point, and proper sign.

Question 19

A temperature function $T$ satisfies $T(10)=68$ and $T'(10)=1.5$ ; find the tangent line at $t=10$ .

$y=68+1.5(t-10)$ (correct answer)
$y=10+1.5(t-68)$
$y=68-1.5(t-10)$
$y=68+1.5t$
$y=1.5+68(t-10)$

Explanation: This problem requires finding the tangent line (linearization) to temperature function T at t = 10. The tangent line formula is y = f(a) + f'(a)(t - a), where we use t instead of x for time. With T(10) = 68 and T'(10) = 1.5, the tangent line is y = 68 + 1.5(t - 10). This line passes through (10, 68) with slope 1.5, meaning temperature increases by 1.5 degrees per unit time near t = 10. Choice D incorrectly writes y = 68 + 1.5t, which equals 68 + 1.5(10) = 83 at t = 10, not the required 68. For linearization: identify the point, evaluate the function and derivative there, then write y = f(a) + f'(a)(variable - a).

Question 20

A volume function $V$ satisfies $V(1)=9$ and $V'(1)=12$ ; which gives the tangent line at $x=1$ ?

$y=9+12(x-1)$ (correct answer)
$y=1+12(x-9)$
$y=9-12(x-1)$
$y=9+12x$
$y=12+9(x-1)$

Explanation: This problem requires finding the tangent line to volume function V at x = 1. The tangent line formula is y = V(a) + V'(a)(x - a) where a is the point of tangency. With V(1) = 9 and V'(1) = 12, the tangent line is y = 9 + 12(x - 1). This line passes through (1, 9) with slope 12, providing the linear approximation to V near x = 1. Choice D incorrectly writes y = 9 + 12x, which equals 9 + 12(1) = 21 at x = 1, not the required 9. Remember: linearization always uses the point-slope form centered at the given point, never forget the (x - a) structure.