Chapter 5, Problem 61EP

To determine

To find: The function $f\left(x\right)$ .

Answer to Problem 61EP

The function f(x) is $4x^3-9x^2+20$ .

Explanation of Solution

  $\begin{array}{l}f\text{'}(x)=a{x}^2+bx\\ f\text{'}(0)=a+b=-6\\ \text{ b = - (a+6) }\mathrm{......}\text{(1)}\end{array}$

  $\begin{array}{l}f\text{'}\text{'}(x)=2\text{a}x+\text{b}\\ \text{ = 2a + b = 6 }\mathrm{......}\text{(2)}\end{array}$

Substitute the equation (1) in equation (2).

  $\begin{array}{l}f\text{'}\text{'}(x)\text{ = 2a + b = -6 }\\ f\text{'}\text{'}(x)=2a-a-6=6\\ \text{ a = 12}\end{array}$

Now,

  $\begin{array}{c}\text{ 2(12) + b = -6 }\\ \text{b = -18}\end{array}$

  $f\text{'}(x)=12x^2-18x$

Integrate the equation to find the function $f\left(x\right)$ :

  $\begin{array}{l}f\text{'}(x)=12x^2-18x\\ f(x)=12.\left(\frac{x^3}{3}\right)-18\left(\frac{x^2}{2}\right)\\ f(x)=4x^3-9x^2+\text{c}\end{array}$

As,

  $\begin{array}{c}{\displaystyle \underset{1}{\overset{2}{\int }}f(x)dx}=14\\ {\displaystyle \underset{1}{\overset{2}{\int }}(4x^3-9x^2+\text{c})dx}=14\\ \left[4\left(\frac{{(2)}^4}{4}\right)-\left(\frac{1}{4}\right)\right]-9\left[\frac{{(2)}^3}{3}-\frac{{(1)}^3}{3}\right]+\text{c}\left[2-1\right]=14\\ 4\times \frac{15}{4}-9\times \frac{7}{3}+\text{c}=14\\ 15-\frac{63}{3}+\text{c}=14\\ \text{c}=20\end{array}$

Hence, the equation is $f(x)=4x^3-9x^2+20$ .

Conclusion:

The function $f\left(x\right)$ is $4x^3-9x^2+20$ .