Chapter 5.5, Problem 10E

a.

To determine

To calculate:

The appropriate value of $f\left(x\right)={\left(4+3x\right)}^{\frac{1}{3}}$

Answer to Problem 10E

The appropriate value of $f\left(x\right)={\left(4+3x\right)}^{\frac{1}{3}}$ is $4+\frac{3x}{3}$

Explanation of Solution

Given information:

The linear approximation of ${\left(1+x\right)}^k\approx 1+kx$

Calculation:

Change the expression to the form ${\left(1+y\right)}^k$ where $k$ is a real number and $y$ is a function of $x$ that is close to zero when $x$ is close to zero. The approximation is then given by $1+ky$

  $\begin{array}{l}f\left(x\right)={\left(4+3x\right)}^{\frac{1}{3}}\\ f\left(x\right)\approx \left\{4+\left(\frac{1}{3}\right)\left(3x\right)\right\}\\ f\left(x\right)=4+\frac{3x}{3}\end{array}$

Therefore, the appropriate value of $f\left(x\right)={\left(4+3x\right)}^{\frac{1}{3}}$ is $4+\frac{3x}{3}$

b.

To determine

To calculate:

The appropriate value of $f\left(x\right)=\sqrt{2+x^2}$

Answer to Problem 10E

The appropriate value of $f\left(x\right)=\sqrt{2+x^2}$ is $2+\frac{1}{2}{x}^2$

Explanation of Solution

Given information:

The linear approximation of ${\left(1+x\right)}^k\approx 1+kx$

Calculation:

Change the expression to the form ${\left(1+y\right)}^k$ where $k$ is a real number and $y$ is a function of $x$ that is close to zero when $x$ is close to zero. The approximation is then given by $1+ky$

  $\begin{array}{l}f\left(x\right)=\sqrt{2+x^2}\\ f\left(x\right)={\left(2+x^2\right)}^{\frac{1}{2}}\\ f\left(x\right)\approx \left\{2+\left(\frac{1}{2}\right)\left(x^2\right)\right\}\\ f\left(x\right)=2+\frac{1}{2}{x}^2\end{array}$

Therefore, the appropriate value of $f\left(x\right)=\sqrt{2+x^2}$ is $2+\frac{1}{2}{x}^2$

c.

To determine

To calculate:

The appropriate value of $f\left(x\right)=\sqrt[3]{{\left(1-\frac{1}{2+x}\right)}^2}$

Answer to Problem 10E

The appropriate value of $f\left(x\right)=\sqrt[3]{{\left(1-\frac{1}{2+x}\right)}^2}$ is $1+\frac{4}{6+x}$

Explanation of Solution

Given information:

The linear approximation of ${\left(1+x\right)}^k\approx 1+kx$

Calculation:

Change the expression to the form ${\left(1+y\right)}^k$ where $k$ is a real number and $y$ is a function of $x$ that is close to zero when $x$ is close to zero. The approximation is then given by $1+ky$

  $\begin{array}{l}f\left(x\right)=\sqrt[3]{{\left(1-\frac{1}{2+x}\right)}^2}\\ f\left(x\right)={\left\{{\left(1-\frac{1}{2+x}\right)}^2\right\}}^{-\frac{2}{3}}\\ f\left(x\right)={\left(1-\frac{1}{2+x}\right)}^{-\frac{4}{3}}\\ f\left(x\right)\approx \left\{1-\left(-\frac{4}{3}\right)\frac{1}{2+x}\right\}\\ f\left(x\right)=1-\left(-\frac{4}{6+x}\right)\\ f\left(x\right)=1+\frac{4}{6+x}\end{array}$

Therefore, the appropriate value of $f\left(x\right)=\sqrt[3]{{\left(1-\frac{1}{2+x}\right)}^2}$ is $1+\frac{4}{6+x}$