Chapter 3.9, Problem 36E

a.

To determine

Use a linear approximation to estimate $g\left(1.95\right)and,g\left(2.05\right).$

Answer to Problem 36E

  $\underset{\_}{g\left(1.95\right)\approx -4.15},\underset{\_}{g\left(2.05\right)\approx -3.85}$

Explanation of Solution

Given information:

Suppose that we don’t have a formula for $g\left(x\right)$ but we know that $g\left(2\right)=-4$ and $g\text{'}\left(x\right)=\sqrt{x^2+5}$ for all $x$ .

Calculation:

Consider the given function,

  $g\left(2\right)=-4$ and $g\text{'}\left(x\right)=\sqrt{x^2+5}$

So,

  $\begin{array}{l}g\text{'}\left(2\right)=\sqrt{2^2+5}\\ g\text{'}\left(2\right)=\sqrt{4+5}=3\end{array}$

The linear approximation at $2$ is,

  $\begin{array}{l}L\left(x\right)=g\left(2\right)+g\text{'}\left(2\right)\left(x-2\right)\\ =-4+3\left(x-2\right)\\ =3x-10\\ g\left(x\right)\approx 3x-\mathrm{10.....}\left(1\right)\\ put,x=1.95,ine{q}^n\left(1\right)\\ g\left(1.95\right)\approx 3\left(1.95\right)-10\\ \underset{\_}{g\left(1.95\right)\approx -4.15}\\ put,x=2.05,ine{q}^n\left(1\right)\\ g\left(2.05\right)\approx 3\left(2.05\right)-10\\ \underset{\_}{g\left(2.05\right)\approx -3.85}\end{array}$

Hence, a linear approximation to estimate $g\left(1.95\right)and,g\left(2.05\right).$ is, $\underset{\_}{g\left(1.95\right)\approx -4.15},\underset{\_}{g\left(2.05\right)\approx -3.85}$

b.

To determine

Estimate that part $\left(a\right)$ is too large or small.

Answer to Problem 36E

  $\text{Too Small}$

Explanation of Solution

Given information:

Suppose that we don’t have a formula for $g\left(x\right)$ but we know that $g\left(2\right)=-4$ and $g\text{'}\left(x\right)=\sqrt{x^2+5}$ for all $x$ .

Calculation:

The graph of the function shows that the tangent line lies below the curve.

  

Hence, approximation is $\text{Too Small}$ .