Chapter 6.3, Problem 4QQ

(a)

To determine

f(x) if the graph of f is tangent to the line at the point [0, -5].

Answer to Problem 4QQ

Function of x:

  $f\left(x\right)=x^3+6x^2+4x-5$

Explanation of Solution

Given information:

f be a function

Such that

  $f\text{'}\text{'}\left(x\right)=6x+12$

f is tangent to the line $4x-y=5$ at point [0, -5].

Tangent line to be rewritten in slope − intercept form:

  $4x-y=5$

Subtract 4x from both sides:

  $-y=5-4x$

Divide both sides by -1:

  $y=4x-5$

Since the graph of f(x) is tangent to the line $y=4x-5$ at [0, -5],

Then

  $f\text{'}\left(0\right)=4$

And

  $f\left(0\right)=5$

Now,

Find $f\text{'}$ by finding the anti-derivative of $f\text{'}\text{'}$ :

  $f\text{'}\left(x\right)={\displaystyle \int f"\left(x\right)}dx={\displaystyle \int \left(6x+12\right)}dx=3x^2+12x+C_1$

Where,

C₁is constant

Now,

Substitute $f\text{'}\left(0\right)=4$ into the first derivative function to find C₁:

  $f\text{'}\left(0\right)=3{\left(0\right)}^2+12\left(0\right)+C_1=4$

Thus,

  $C_1=4$

Then

The first derivative becomes

  $f\text{'}\left(x\right)=3x^2+12x+4$

Now,

Find f by finding the anti-derivative of $f\text{'}$ :

  $f\left(x\right)={\displaystyle \int f\text{'}\left(x\right)}dx={\displaystyle \int \left(3x^2+12x+4\right)}dx=x^3+6x^2+4x+C_2$

Where,

C₂is constant

Substitute $f\left(0\right)=-5$ into f to find C₂:

  $f\left(0\right)={\left(0\right)}^3+6{\left(0\right)}^2+4\left(0\right)+C_2=-5$

Thus,

  $C_2=-5$

Therefore,

The function becomes

  $f\left(x\right)=x^3+6x^2+4x-5$

(b)

To determine

Average value of f(x) on the closed interval [-1, 1].

Answer to Problem 4QQ

Average value of f(x):

  $\overline{f}=-3$

Explanation of Solution

Given information:

f be a function

Such that

  $f\text{'}\text{'}\left(x\right)=6x+12$

On closed interval [-1, 1]:

Average value,

  $\begin{array}{l}\overline{f}=\frac{1}{b-a}{\displaystyle {\int }_{-1}^1\left(x^3+6x^2+4x-5\right)}\\ =\frac{1}{1-\left(-1\right)}\left[{\frac{x^4}{4}+2x^3+2x^2-5x|}_{-1}^1\right]\\ =\frac{1}{2}\left[0.25+2+2-5-\left(0.25-2+2+5\right)\right]\\ =\frac{1}{2}\left[-0.75-5.25\right]\\ =\frac{1}{2}\left[-6\right]\\ =-3\end{array}$