Chapter 4.3, Problem 28E

(a)

To determine

To Find: The value of $f(1)\text{and}f\text{'}(1)$ of the given function.

Answer to Problem 28E

  $f(1)=3\text{and}f\text{'}(1)=12$

Explanation of Solution

Given information: Given function $f(x)=x^5+2x^3+x-1$

Formula used: $\frac{d(x^n)}{dx}=n{x}^{n-1}\text{and}\frac{d(\text{constant})}{dx}=0$ with addition and subtraction rule.

Calculation:

  $f(x)=x^5+2x^3+x-1$

Differentiate with respect to x

  $\begin{array}{l}f\text{'}(x)=\frac{d(x^5)}{dx}+2\times \frac{d(x^3)}{dx}+\frac{d(x)}{dx}-\frac{d(1)}{dx}\\ f\text{'}(x)=5x^4+2\times 3x^2+1-0\\ f\text{'}(x)=5x^4+6x^2+1\end{array}$

Now,

  $\begin{array}{l}f(x)=x^5+2x^3+x-1\\ f(1)=1^5+2\times 1^3+1-1\\ f(1)=1+2+0\\ f(1)=3\\ also,\\ f\text{'}(x)=5x^4+6x^2+1\\ f\text{'}(1)=5\times 1^4+6\times 1^2+1\\ f\text{'}(1)=12\end{array}$

(b)

To determine

To Find: Value of $f^{-1}(3)\text{and}(f^{-1})\text{'}(3)$ of the given function.

Answer to Problem 28E

  $f^{-1}(3)=1\text{and}(f^{-1})\text{'}(3)=\frac{1}{12}$

Explanation of Solution

Given information: Given function $f(x)=x^5+2x^3+x-1$

Formula used: $\frac{d(x^n)}{dx}=n{x}^{n-1}\text{,}\frac{d(\text{constant})}{dx}=0\text{and}(f^{-1})\text{'}(x)=\frac{1}{f\text{'}(f^{-1}(x))}$ with addition and subtraction rule.

Calculation:

  $f(x)=x^5+2x^3+x-1$

Differentiate with respect to x

  $\begin{array}{l}f\text{'}(x)=\frac{d(x^5)}{dx}+2\times \frac{d(x^3)}{dx}+\frac{d(x)}{dx}-\frac{d(1)}{dx}\\ f\text{'}(x)=5x^4+2\times 3x^2+1-0\\ f\text{'}(x)=5x^4+6x^2+1\\ f\text{'}(x)>0\end{array}$

Now,

  $\begin{array}{l}f(1)=1^5+2\times 1^3+1-1\\ f(1)=3\\ f^{-1}(f(1))=f^{-1}(3)\\ \text{Thus,}{f}^{-1}(3)=1\end{array}$

Also,

  $\begin{array}{l}(f^{-1})\text{'}(3)=\frac{1}{f\text{'}(f^{-1}(3))}\\ (f^{-1})\text{'}(3)=\frac{1}{f\text{'}(1)}\\ (f^{-1})\text{'}(3)=\frac{1}{5\times 1^4+6\times 1^2+1}\\ (f^{-1})\text{'}(3)=\frac{1}{12}\end{array}$