Chapter 5, Problem 39RE

a.

To determine

To find the linearization of $f$ at $x=0$

Answer to Problem 39RE

The Linearization of function is $L\left(x\right)=-1$

Explanation of Solution

Given:

Let $f$ be a function with $f^{\prime }\left(x\right)=\sin x^2$ and $f\left(0\right)=-1$

Calculation:

If $f$ is a differentiable at $x=a$ , then the equation of the tangent line,

  $L\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$

Defines the linearization of $f$ at $a$ .

Since, $f\left(0\right)=-1$ , the point of tangency is $\left(0,-1\right)$ .also, $f^{\prime }\left(x\right)=\sin x^2$ , the slope of the tangent line is $f^{\prime }\left(0\right)=\sin {\left(0\right)}^2=0$

So,

  $\begin{array}{l}L\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)\\ L\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)\left(x-0\right)\\ L\left(x\right)=-1+\left(0\right)\left(x-0\right)\\ L\left(x\right)=-1\end{array}$

Therefore, the Linearization of function is $L\left(x\right)=-1$ .

b.

To determine

To find the approximate value of $f$ at $x=0.1$

Answer to Problem 39RE

The approximate value of $f$ at $x=0.1$ is $-1$ .

Explanation of Solution

Given:

Let $f$ be a function with $f^{\prime }\left(x\right)=\sin x^2$ and $f\left(0\right)=-1$

Calculation:

If $f$ is a differentiable at $x=a$ , then the equation of the tangent line,

  $L\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$

Defines the linearization of $f$ at $a$ . The approximation $f\left(x\right)\approx L\left(x\right)$ is the standard linear approximation of $f$ at $a$ .

Since , the Linearization of $f$ is $L\left(x\right)=-1$ it means that for all values of $x$ , $f$ is approximately $-1$ .

Therefore, the approximate value of $f$ at $x=0.1$ is $-1$ .

c.

To determine

To find whether the actual value of $f$ at $x=0.1$ is greater than or less than the approximation in (b).

Answer to Problem 39RE

The approximate value is less than the actual value.

Explanation of Solution

Given:

Let $f$ be a function with $f^{\prime }\left(x\right)=\sin x^2$ and $f\left(0\right)=-1$

Calculation:

Since, the linearization of $f$ is $L\left(x\right)=-1$ which was centered at $x=0$

Now $f^{″}\left(x\right)=2x\cos x^2$ and for $x>0$ , $f^{″}\left(x\right)>0$ means the curve is concave up.

Since, linearization provides tangent which is a line , so approximate value is less than the actual value.