Question

The function f(x) = - 4x ^ 3 + 27.24x ^ 2 + 220.4352x - 4.76 is increasing on the open interval

Answer 1

Given:

The function is:

$f(x)=-4x^3+27.24x^2+220.4352x-4.76$

Find:

(a)

The increase in the open interval

(b)

The decrease in the open interval

(c)

The local maximum

Explanation-:

The function is:

$f(x)=-4x^3+27.24x^2+220.4352x-4.76$

The first derivative of a function is:

$f^{\prime}(x)=-12x^2+54.58x+220.4352$

For increasing at x

$\begin{gathered} f^{\prime}(x)>0 \\ \\ f(x)\text{ Increasing at }x \end{gathered}$

For decreasing at x

$\begin{gathered} f^{\prime}(x)<0 \\ \\ f(x)\text{ is decreasing at }x \end{gathered}$

f'(x) is zero at a critical point:

$\begin{gathered} f^{\prime}(x)=0 \\ \\ -12x^2+54.58x+220.4352=0 \\ \\ x=-\frac{129}{50}\text{ and }x=\frac{178}{25} \end{gathered}$

So, the interval of increase is:

$\text{ Increasing open interval }=(-\frac{129}{50},\frac{178}{25})$

The decreasing interval is all value - increasing interval.

So,

Decreasing interval is:

$=(-\infty,-\frac{129}{50})\cup(\frac{178}{25},\infty)$

The function local maxima is:

The local maxima

$\begin{gathered} f^{\prime}(x)=0 \\ \\ \text{ At }x=-\frac{129}{50}\text{ and }x=\frac{178}{25} \\ \end{gathered}$

For local maxima check the function value:

$\begin{gathered} f(x)=f(-\frac{129}{50}) \\ \\ =-4(-\frac{129}{50})^3+27.24(\frac{-129}{50})^2+220.4352(-\frac{129}{50})-4.76 \\ \\ =-323.468 \end{gathered}$

Check the value of another critical point:

$\begin{gathered} f(x)=f(\frac{178}{25}) \\ \\ =-4(\frac{178}{25})^3+27.24(\frac{178}{25})^2+220.4352(\frac{178}{25})-4.76 \\ \\ =1501.8776 \end{gathered}$

So, the local maxima at

$x=\frac{178}{25}$

The local maxima is 178/25