# AP Calculus AB: 4.6 Approximating Values of a Function Using Local Linearity and Linearization – Exam Style questions with Answer- MCQ

### Question The graph of f′, the derivative of the function f, is shown above. If f(4)=−1, what is the approximation for f(4.5) using the line tangent to the graph of f at x=4 ?

A -4

B -1

C 2

D  6

Ans:C

An equation of the tangent line at x=a is y=f(a)+f′(a)(x−a). Here a=4, so f(a)=−1 and f′(a)=6. The value of y when x=4.5 would be an approximation to f(4.5), as follows.
f(4.5)≈f(4)+6(4.5−4)=−1+3=2

### Question

For the function f, $$f'(x)=2x+1$$ and f(1)=4.What is the approximation for f(1,2) found by using the line tangent to the graph of f at  x=1?

A 0.6

B 3.4

C 4.2

D 4.6

E 4.64

Ans:D

### Question

Let f be the function given by f(x) = 2 cos x + 1. What is the approximation for f(1.5) found by using the line tangent to the graph of f at $$x=\frac{\pi }{2}$$ ?

A -2

B 1

C $$\pi -2$$

D $$4-\pi$$

Ans:C

### Question

The function $$f$$ is twice differentiable with $$f(2)=1, f^{\prime}(2)=4$$, and $$f^{”}(2)=3$$. What is the value of the approximation of $$f(1.9)$$ using the line tangent to the graph of $$f$$ at $$x=2$$ ?

A 0.4

B 0.6

C 0.7

D 1.3

E 1.4

Ans:B

$f(2)=1, f^{\prime}(2)=4 f^{\prime \prime}(2)=3 x=2 f(1.9)=?$

The formula for the slope of a tangent line to a curve $$y=f(x)$$ at the point $$(a, f(a))$$ is:
\begin{aligned} m & =\frac{y-f(a)}{x-a} \\ f^{\prime}(a) & =\frac{y-f(a)}{x-a} \end{aligned}

Rewrite the above expression for the variable $$y$$ or the function $$f(x)$$.
\begin{aligned} (x-a) f^{\prime}(a) & =y-f(a) \\ y-f(a) & =(x-a) f^{\prime}(a) \\ y & =(x-a) f^{\prime}(a)+f(a) \\ f(x) & =(x-a) f^{\prime}(a)+f(a) \quad[\because y=f(x)] \end{aligned}

Substitute $$a=2$$ in the above function.
$\begin{array}{rlr} f(x) & =(x-2) f^{\prime}(2)+f(2) & \\ & =(x-2) 4+1 \quad\left[\because f^{\prime}(2)=4 \text { and } f(2)=1\right] \\ & =4 x-8+1 \\ & =4 x-7 \end{array}$

Substitute $$x=1.9$$ in the above function and solve for
\begin{aligned} & f(1.9) . \\ & \begin{aligned} f(1.9) & =4(1.9)-7 \\ & =7.6-7 \\ & =0.6 \end{aligned} \end{aligned}

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