Chapter 9.3, Problem 30E

$\text{(a)}$

To determine

To find: The linearization.

Answer to Problem 30E

The answer is:

At $x=0$ the linearization of $f$ is $f(x)\approx 1$ .

Explanation of Solution

Given information:

The function is $f(x)=\sec x$ .

Calculation:

The function's first two derivatives are determined

  $\begin{array}{c}{f}^{\prime }(x)=\frac{\sin x}{{\cos }^{2}x}\\ f^{″}(x)=\frac{{\cos }^{3}x+2{\sin }^{2}x\cos x}{{\cos }^{4}x}\\ =\frac{1+{\sin }^{2}x}{{\cos }^{2}x}\end{array}$

The linearization of $f$ at $x=0$

Is:

  $\begin{array}{c}f(x)\approx P_1(x)\\ =\sec 0+\frac{\sin 0}{{\cos }^{2}0}x\\ =1\end{array}$

Therefore, the required at $x=0$ the linearization of $f$ is $f(x)\approx 1$ .

$\text{(b)}$

To determine

To find: The quadratic approximation.

Answer to Problem 30E

At $x=0$ , the quadratic approximation is $f(x)\approx 1+x^2/2$ .

Explanation of Solution

Given information:

The function is $f(x)=\sec x$

Calculation:

At $x=0$ , the quadratic approximation of $f$ at $x=0$

Is:

  $\begin{array}{c}f(x)\approx P_2(x)\\ =P_1(x)+\frac{\frac{1+{\sin }^{2}0}{{\cos }^{2}0}}{2!}{x}^2\\ =1+\frac{x^2}{2}\end{array}$

The graph of $f$ and the graphs of the approximations are shown in the figure below, and they behave at $x=0$ as would have predicted.

  

Therefore, the required at $x=0$ , the quadratic approximation is $f(x)\approx 1+x^2/2$ .