Chapter 6, Problem 59RE

To determine

To calculate:The value of f(x)

Answer to Problem 59RE

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

Explanation of Solution

Given information:

Let/be a differentiable function with the following properties.

  $\begin{array}{l}i)f\text{'}\left(x\right)=a{x}^2+bx\\ ii)f\text{'}\left(1\right)=-6\text{and}f\text{'}\text{'}\left(x\right)=6\\ iii){\displaystyle \underset{1}{\overset{2}{\int }}f\left(x\right)}dx=14\end{array}$

Formula used:

  $f\text{'}\left(x\right)=a{x}^2+bx$

Calculation:

Since it has been given that

  $f\text{'}\left(x\right)=a{x}^2+bx$

Hence,

  $\begin{array}{l}{f}^{\text{'}}\left(1\right)=a{\left(1\right)}^2+b\left(1\right)\\ -6=a+b\end{array}$

Now take the derivative of $f\text{'}\left(x\right)$

  $\begin{array}{l}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(a{x}^2+bx\right)\\ =\frac{d\left(a{x}^2\right)}{dx}+\frac{d\left(b{x}^2\right)}{dx}\\ f\text{'}\text{'}\left(x\right)=a\left(2x\right)+b\end{array}$

Now,

  $\begin{array}{l}f\text{'}\text{'}\left(1\right)=6\\ 2a\left(1\right)+b=6\\ 2a+b=6\end{array}$

Solve the equations for a and b

  $\begin{array}{l}a+b=-6\\ 2a+b=6\end{array}$

Then,

  $\left(a,b\right)\equiv \left(12,-18\right)$

Hence,

  $f\text{'}\left(x\right)=12x^2-18x$

Integrate both sides

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

Where c is integration constant

Now,

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{2}{\int }}f\left(x\right)}dx=14\\ {\displaystyle \underset{1}{\overset{2}{\int }}\left(4x^3-9x^2+c\right)dx=14}\\ {\left[x^4-3x^3+cx\right]}_1^2=14\\ \left[2^4-3{\left(2\right)}^3+2c-1^4+3{\left(1\right)}^3-c\left(1\right)\right]=14\\ c-6=14\\ c=20\end{array}$

Now, substitute this value of c in equation f(x)

  $f\left(x\right)=4x^3-9x^2+20$

Conclusion:

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