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

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

▶️Answer/Explanation

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

▶️Answer/Explanation

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 \(4-\pi \)

▶️Answer/Explanation

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

▶️Answer/Explanation

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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