Local Linearity and Linearization - AP Calculus BC

Let $h$ be differentiable with $h(-1)=3$ and $h'(-1)=0$ . What is the linearization at $x=-1$ ?
Answer: $L(x)=3$ 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.
Given $H(4)=\!-8$ and $H'(4)=\!-\tfrac{3}{2}$ , which is the tangent line at $x=4$ ?
Answer: $y=-8-\tfrac{3}{2}(x-4)$ This problem involves linearization with negative function value and negative fractional derivative. Using y = H(a) + H'(a)(x - a) with a = 4, H(4) = -8, and H'(4) = -3/2. The tangent line becomes y = -8 + (-3/2)(x - 4) = -8 - (3/2)(x - 4). The line passes through (4, -8) with slope -3/2, decreasing by 3 units for every 2 units increase in x. Both negative values reinforce the decreasing trend from an already negative starting point. Choice C uses +3/2 instead of -3/2, representing increasing behavior when both values suggest continued decrease. Fractional negative derivatives require careful sign tracking throughout the calculation.
For $f$ , $f(6)=2$ and $f'(6)=\tfrac34$ . Which is the tangent line at $x=6$ ?
Answer: $y=2+\tfrac34(x-6)$ This question asks for the tangent line with a fractional derivative. Using y = f(a) + f'(a)(x - a) with a = 6, f(6) = 2, and f'(6) = 3/4. The tangent line equation becomes y = 2 + (3/4)(x - 6). This line passes through the point (6, 2) and has slope 3/4, representing a moderate positive increase. The fractional slope means the function increases by 3 units for every 4 units of increase in x. Choice C uses -3/4 instead of +3/4, which would make the function decreasing rather than increasing. When working with fractional derivatives, maintain the exact fraction and verify the sign corresponds to increasing or decreasing behavior.
A position function $s$ has $s(3)=20$ and $s'(3)=6$ . Which linearization approximates $s(x)$ near $x=3$ ?
Answer: $L(x)=20+6(x-3)$ 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.
Given $p(4)=7$ and $p'(4)=2$ , which is the tangent line to $p$ at $x=4$ ?
Answer: $y=7+2(x-4)$ 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).
A volume function $V$ satisfies $V(1)=9$ and $V'(1)=12$ ; which gives the tangent line at $x=1$ ?
Answer: $y=9+12(x-1)$ 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.
Let $k(7)=\tfrac{9}{2}$ and $k'(7)=-\tfrac{1}{3}$ . Which is the tangent line at $x=7$ ?
Answer: $y=\tfrac{9}{2}-\tfrac{1}{3}(x-7)$ 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.
A function $S$ has $S(11)=\!-2$ and $S'(11)=9$ . Which is the tangent line at $x=11$ ?
Answer: $y=-2+9(x-11)$ 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.
Let $q$ be differentiable with $q(-2)=4$ and $q'(-2)=-5$ . Find the linearization at $x=-2$ .
Answer: $L(x)=4-5(x+2)$ This problem involves linearization at a negative x-value with specific attention to signs. Using L(x) = q(a) + q'(a)(x - a) with a = -2, q(-2) = 4, and q'(-2) = -5. The linearization becomes L(x) = 4 + (-5)(x - (-2)) = 4 - 5(x + 2). Note that (x - (-2)) simplifies to (x + 2), which is crucial for the correct form. The tangent line passes through (-2, 4) with slope -5, decreasing rapidly as x moves away from -2. Choice C uses +5 instead of -5, giving an incorrect positive slope for a decreasing function. When linearizing at negative points, carefully handle the signs in both the derivative and the (x - a) expression.
If $f(2)=\!-\tfrac{3}{2}$ and $f'(2)=\tfrac{5}{4}$ , which is the linearization at $x=2$ ?
Answer: $L(x)=-\tfrac{3}{2}+\tfrac{5}{4}(x-2)$ This problem involves linearization with negative fractional function value and fractional derivative. Using L(x) = f(a) + f'(a)(x - a) with a = 2, f(2) = -3/2, and f'(2) = 5/4. The linearization becomes L(x) = -3/2 + (5/4)(x - 2). The tangent line passes through (2, -3/2) with slope 5/4, increasing by 5 units for every 4 units increase in x. Despite the negative starting point, the positive fractional derivative indicates upward movement. Choice C uses -5/4 instead of +5/4, representing continued decrease rather than recovery. Mixed fractional values require careful attention to signs and proper algebraic combination.
A differentiable function $B$ satisfies $B(-6)=5$ and $B'(-6)=\tfrac{2}{5}$ . Find the tangent line at $x=-6$ .
Answer: $y=5+\tfrac{2}{5}(x+6)$ This linearization involves a negative point with positive function value and fractional derivative. Using $y = B(a) + B'(a)(x - a)$ with $a = -6$ , $B(-6) = 5$ , and $B'(-6) = \frac{2}{5}$ . The tangent line becomes $y = 5 + \frac{2}{5}(x - (-6)) = 5 + \frac{2}{5}(x + 6)$ . The line passes through $(-6, 5)$ with slope $\frac{2}{5}$ , increasing slowly by 2 units for every 5 units increase in x. The positive function value and positive derivative both contribute to upward trending behavior. Choice C uses $-\frac{2}{5}$ instead of $+\frac{2}{5}$ , representing decreasing behavior when both values indicate increase. Fractional derivatives require precision in sign determination and algebraic manipulation.
Let $f(\!-\tfrac12)=\!-2$ and $f'(\!-\tfrac12)=3$ . What is the linearization at $x=-\tfrac12$ ?
Answer: $L(x)=-2+3\left(x+\tfrac12\right)$ This problem involves linearization at a negative fractional point with negative function value and positive derivative. Using L(x) = f(a) + f'(a)(x - a) with a = -1/2, f(-1/2) = -2, and f'(-1/2) = 3. The linearization becomes L(x) = -2 + 3(x - (-1/2)) = -2 + 3(x + 1/2). The tangent line passes through (-1/2, -2) with slope 3, increasing rapidly by 3 units per unit increase in x. The positive derivative overcomes the negative starting point, showing strong recovery. Choice C uses -3 instead of +3, representing continued decrease rather than the intended increase. When negative function values combine with positive derivatives, linearization shows upward recovery trends.
Given $g(1)=-2$ and $g'(1)=4$ , which equation gives the linear approximation to $g(x)$ near $x=1$ ?
Answer: $L(x)=-2+4(x-1)$ This problem asks for the linear approximation to g(x) near x = 1. The linearization formula is L(x) = g(a) + g'(a)(x - a) where a is the approximation point. Given g(1) = -2 and g'(1) = 4, we have L(x) = -2 + 4(x - 1). This tangent line passes through (1, -2) with slope 4, providing the best linear approximation to g near x = 1. Choice C incorrectly writes L(x) = -2 + 4x, which equals -2 + 4(1) = 2 at x = 1, not the required -2. Always use the point-slope form: start with the function value at the point, then add the derivative times (x - point).
Given $A(3)=\tfrac{1}{4}$ and $A'(3)=\!-8$ , what is the linear approximation near $x=3$ ?
Answer: $L(x)=\tfrac14-8(x-3)$ This problem involves linearization with fractional function value and negative integer derivative. Using L(x) = A(a) + A'(a)(x - a) with a = 3, A(3) = 1/4, and A'(3) = -8. The linear approximation becomes L(x) = 1/4 + (-8)(x - 3) = 1/4 - 8(x - 3). The tangent line passes through (3, 1/4) with slope -8, decreasing very rapidly from the small positive starting value. The large negative derivative dominates the behavior, creating steep decrease. Choice C uses +8 instead of -8, representing increasing behavior when the derivative indicates rapid decrease. Large negative derivatives create steep decreasing linearizations regardless of function value magnitude.
If $w(2)=-3$ and $w'(2)=-4$ , what is the tangent line equation at $x=2$ ?
Answer: $y=-3-4(x-2)$ This linearization involves negative values for both function value and derivative. Using y = w(a) + w'(a)(x - a) with a = 2, w(2) = -3, and w'(2) = -4. The tangent line becomes y = -3 + (-4)(x - 2) = -3 - 4(x - 2). Both the function value and derivative are negative, creating a tangent line that passes through (2, -3) and decreases rapidly with slope -4. The linearization shows the function decreasing by 4 units per unit increase in x near x = 2. Choice C uses +4 instead of -4, suggesting increasing behavior when the function is actually decreasing. Always verify that negative derivatives produce decreasing tangent lines.
For differentiable $h$ , if $h(-1)=3$ and $h'(-1)=-2$ , find the tangent line to $y=h(x)$ at $x=-1$ .
Answer: $y=3-2(x+1)$ This problem requires finding the tangent line to y = h(x) at x = -1. The tangent line formula is y = h(a) + h'(a)(x - a) where a is the point of tangency. Given h(-1) = 3 and h'(-1) = -2, the tangent line is y = 3 + (-2)(x - (-1)) = 3 - 2(x + 1). This line passes through (-1, 3) with slope -2, providing the linear approximation to h near x = -1. Choice B incorrectly writes y = 3 - 2(x - 1), which equals 3 - 2(-1 - 1) = 7 at x = -1, not the required 3. For linearization at negative values: be careful with signs when substituting x - a where a is negative.
A mass function $M$ satisfies $M(2)=9$ and $M'(2)=-1$ . Which linearization approximates $M(t)$ near $t=2$ ?
Answer: $L(t)=9-1(t-2)$ This problem involves linearization of a mass function with negative derivative. Using L(t) = M(a) + M'(a)(t - a) with a = 2, M(2) = 9, and M'(2) = -1. The linear approximation becomes L(t) = 9 + (-1)(t - 2) = 9 - 1(t - 2). The tangent line starts at (2, 9) and decreases with slope -1, meaning mass decreases by 1 unit per unit increase in time. This linearization provides the best linear estimate of the mass function near t = 2. Choice C uses +1 instead of -1, suggesting increasing mass rather than decreasing. For physical applications, ensure the derivative sign matches the expected behavior (increasing or decreasing quantity).
If $c(-3)=10$ and $c'(-3)=7$ , what is the tangent line to $c$ at $x=-3$ ?
Answer: $y=10+7(x+3)$ 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.
A function $M$ has $M(\pi)=2\pi$ and $M'(\pi)=2$ . What is the linear approximation near $x=\pi$ ?
Answer: $L(x)=2\pi+2(x-\pi)$ 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.
Given $R(\!-1)=1$ and $R'(\!-1)=\!-\tfrac{7}{3}$ , which is the tangent line at $x=-1$ ?
Answer: $y=1-\tfrac{7}{3}(x+1)$ This problem involves linearization at a negative point with positive function value and negative fractional derivative. Using y = R(a) + R'(a)(x - a) with a = -1, R(-1) = 1, and R'(-1) = -7/3. The tangent line becomes y = 1 + (-7/3)(x - (-1)) = 1 - (7/3)(x + 1). The line passes through (-1, 1) with slope -7/3, decreasing by 7 units for every 3 units increase in x. The positive function value contrasts with the negative derivative, showing initial positive value but decreasing trend. Choice C uses +7/3 instead of -7/3, representing increasing behavior when the derivative indicates decrease. Negative fractional derivatives create specific decreasing rates that must be preserved accurately.

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