Chapter 4.3, Problem 72E

To determine

To find: The value of c for which the given function increases.

Answer to Problem 72E

The value of c for which the given function increases is $\frac{1}{8}<c$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=cx+\frac{1}{x^2+3}$ .

Calculation:

Consider the equation.

  $f\left(x\right)=cx+\frac{1}{x^2+3}$

Differentiate above equation.

  $f^{\prime }\left(x\right)=c-\frac{2x}{{\left(x^2+3\right)}^2}$

The function will be increasing if $0<f\left(x\right)$

  $\begin{array}{l}0<c-\frac{2x}{{\left(x^2+3\right)}^2}\\ \frac{2x}{{\left(x^2+3\right)}^2}<c\end{array}$

Thus, c must be greater than the value of $\frac{2x}{{\left(x^2+3\right)}^2}$

To maximize the expression take the derivative and equate to zero.

  $\begin{array}{l}\frac{dy}{dx}=\frac{2}{{\left(x^2+3\right)}^2}-\frac{8x^2}{{\left(x^2+3\right)}^3}\\ 0=\frac{6x^2-6}{{\left(x^2+3\right)}^3}\\ x=\pm 1\end{array}$

At $x=0$ ,

  $\begin{array}{l}\frac{dy}{dx}=-\frac{\left(-6\right)}{{\left(3\right)}^3}\\ =\frac{2}{9}\end{array}$

Since, the value is positive, it means the right hand point is maximum.

Thus, $\frac{2x}{{\left(x^2+3\right)}^2}$ occurs at $x=1$ .

Substitute $x=1$ in inequality.

  $\begin{array}{l}\frac{2\left(1\right)}{{\left(1^2+3\right)}^2}<c\\ \frac{1}{8}<c\end{array}$