Chapter 4.5, Problem 41E

a.

To determine

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

Answer to Problem 41E

The linearization of $f$at $x=0$is $x+1$ .

Explanation of Solution

Given: The function $f\text{'}(x)=\cos (x^2)$ .

Formula used: $L(x)=f(a)+f\text{'}(a)(x-a)$.

Calculation:

The given function is $f\text{'}(x)=\cos (x^2)$ .

Substitute the value of x equal to $0$ in the equation $f\text{'}(x)=\cos (x^2)$ .

  $\begin{array}{l}f\text{'}(0)=\cos (0^2)\\ f\text{'}(0)=1\end{array}$

The equation of the tangent line $L(x)=f(a)+f\text{'}(a)(x-a)$when it is differentiable at $x=a$ .

It defines the linearization of $f$ at $a$ . Approximation $f(x)\simeq L(x)$ is the standard linear approximation of $f$ at $a$ . The point $x=a$ is the centre of the approximation.

Substitute in the above equation,

  $\begin{array}{l}L(x)=f(0)+f\text{'}(0)(x-0)\\ L(x)=1+1(x)\\ L(x)=x+1\end{array}$

Conclusion:

The linearization of $f$at $x=0$is $x+1$ .

b.

To determine

To find: To find the value of $f$at $x=0.1$.

Answer to Problem 41E

The value of $f$at $x=0.1$is $1.1$ .

Explanation of Solution

Given: The function $f\text{'}(x)=\cos (x^2)$ .

Formula used: $L(x)=f(a)+f\text{'}(a)(x-a)$

Calculation:

The given function is $f\text{'}(x)=\cos (x^2)$ .

Substitute the value of x equal to $0$ in the equation $f\text{'}(x)=\cos (x^2)$ .

  $\begin{array}{l}f\text{'}(0)=\cos (0^2)\\ f\text{'}(0)=1\end{array}$

The equation of the tangent line $L(x)=f(a)+f\text{'}(a)(x-a)$when it is differentiable at $x=a$ .

It defines the linearization of $f$ at $a$ . Approximation $f(x)\simeq L(x)$ is the standard linear approximation of $f$ at $a$ . The point $x=a$ is the centre of the approximation.

Substitute in the above equation,

  $\begin{array}{l}L(x)=f(0)+f\text{'}(0)(x-0)\\ L(x)=1+1(x)\\ L(x)=x+1\end{array}$

Substituting $x=0.1$,$\begin{array}{l}f(x)\simeq x+1\\ f(0.1)\simeq 0.1+1\\ f(0.1)\simeq 1.1\end{array}$

Conclusion:

To find the value of $f$at $x=0.1$is $1.1$ .

c.

To determine

To find: To compare the actual value and estimated value.

Answer to Problem 41E

Actual value is less than the value estimated in part b.

Explanation of Solution

Given: The function $f\text{'}(x)=\cos (x^2)$

Calculation:

If graph is concave up, then linearixation is below the curve. In this case, actual value is greater than the value estimated by linearization. For this, function must meet the condition that

  $f\text{'}\text{'}(x)>0$

If graph is concave down, then linearixation is above the curve. In this case, actual value is less than the value estimated by linearization. For this, function must meet the condition that

  $f\text{'}\text{'}(x)<0$

For the linearization is coincide over the curve, function must meet the condition that

  $f\text{'}\text{'}(x)=0$

Since first derivative is decreasing over the interval $[0,0.1]$ .Thus, second derivatives is negative. So, the graph of function is concave down, Hence, linearization is above the curve. Thus, actual value less than the value estimated in part b.

Conclusion:

Actual value is less than the value estimated in part b.