# The polynomial function is defined by f(x) = 4x^4 + 2x^3 - 8x^2 - 5x + 2. Use a calculator to...

## Question:

The polynomial function is defined by {eq}f(x) = 4x^4 + 2x^3 - 8x^2 - 5x + 2. {/eq} Use a calculator to find all the points (x,f(x)) where there is a local maximum. Round to the nearest hundredth.

## Local maximum

• Local maximum occurs when the value of the function at any point is greater than any other point in the interval.
• In any interval {eq}f(a)>f(x) {/eq}.
• The mathematical condition for local maximum is {eq}f^{\prime}(x)=0 {/eq} and {eq}f^{\prime \prime}(x) < 0 {/eq} .

Given :

{eq}\begin{align} f(x) &= 4x^4 + 2x^3 - 8x^2 - 5x + 2 \\ f^{\prime}(x)&=4(4x^3)+2(3x^2)-8(2x)-5(1)+0 \\ \implies f^{\prime}(x)&=16x^3+6x^2-16x-5\\ f^{\prime \prime}(x)&=16(3x^2)+6(2x)-16(1)-0 \\ \implies f^{\prime \prime}(x)&=48x^2+12x-16 \\ \end{align} {/eq}

For maxima or minima : {eq}f^{\prime}(x)=0 {/eq}

{eq}\implies f^{\prime}(x)=16x^3+6x^2-16x-5 =0 \\ \therefore x=.9766 , -.3060 ,-1.0456 {/eq}

Applying Double Derivative test to identify the points of maximum. At local maxima {eq}f^{\prime \prime}(x) < 0 {/eq}

• {eq}\text{ For } x=.9766 {/eq}

{eq}f^{\prime \prime}(.9766)=41.4991 >0 {/eq}

Hence {eq}x=.9766 {/eq} is the point of local minimum.

• {eq}\text{ For } x=-.3060 {/eq}

{eq}f^{\prime \prime}(-.3060)= -15.1774<0 {/eq}

Hence {eq}x=-.3060 {/eq} is the point of local maximum.

• {eq}\text{ For } x=-1.0456 {/eq}

{eq}f^{\prime \prime}(-1.0456)=23.9302 > 0 {/eq}

Hence {eq}x=-1.0456 {/eq} is the point of local maximum.

Therefore {eq}(-.3060 , 2.7587) {/eq} is the point of local maximum on the curve {eq}f(x) = 4x^4 + 2x^3 - 8x^2 - 5x + 2. {/eq}.