Question

How do I solve and what is the question asking for

Answer 1

To obtain the slope of the tangent line to f(x) at x=2, we need to find f'(x) as shown below

$f^{\prime}(x)=2x-2$

Then, the slope of the tangent line we are looking for is

$f^{\prime}(2)=2\cdot2-2=2=\frac{2}{1}$

Remember that the slope of a line can be interpreted as shown below

$m=\frac{a}{b}\to\text{move a units on the y-ax is direction per each b units on x-axis direction}$

Therefore, we can estimate the value of f(x) using the slope and f(2).

a) x=2.4

$f(2.4)=3.96$

And the estimation using f(2) and f'(2) is

$\begin{gathered} f(2.4)\approx f(2)+f^{\prime}(2)\cdot(2.4-2)=3+2(0.4)=3.8 \\ \end{gathered}$

Then, the exact value at x=2.4 is f(2.4)=3.96, and the approximated value is 3.8

$3.96-3.8=0.16$

We need to repeat these steps with the remaining options.

b) x=2.5

$\begin{gathered} x=2.5 \\ \Rightarrow f(2.5)=4.25 \\ \text{and} \\ f(2.5)\approx\approx f(2)+f^{\prime}(2)\cdot(2.5-2)=3+2(0.5)=4 \\ \Rightarrow4.25-4=0.25$

c) x=2.6

$\begin{gathered} x=2.6 \\ \Rightarrow f(2.6)=4.56 \\ \text{and} \\ f(2.6)\approx f(2)+f^{\prime}(2)\cdot(2.6-2)=3+2(0.6)=4.2 \\ \Rightarrow4.56-4.2=0.36$

d) x=2.7

$\begin{gathered} x=2.7 \\ \Rightarrow f(2.7)=4.89 \\ \text{and} \\ f(2.7)\approx f(2)+f^{\prime}(2)\cdot(2.7-2)=3+2(0.7)=4.4 \\ \Rightarrow4.89-4.4=0.49$

Then, the answer is option d. x=2.7

e) x=2.8

$\begin{gathered} x=2.8 \\ \Rightarrow f(2.8)=5.24 \\ \text{and} \\ \end{gathered}$