Chapter 6.4, Problem 26PPS

To determine

To find: The x coordinate of the relative maxima, minima and graph the function using graphing calculator.

Answer to Problem 26PPS

The x coordinate of the Maxima= $-1.871$ and the $x$ coordinate of minima point is $1.518$ .

Explanation of Solution

Given:

  $f\left(x\right)=x^5-4x^3+3x^2-8x-6$ .

Concept used:

The graph of the function $\text{f}$ is the set of all point in the plane of the form $\left(x,f\left(x\right)\right)$ .

The graph can be defined by the graph of $\text{f}$ and the equation for the graph will be $\text{y=f}\left(\text{x}\right)$ .

The graph of the function is special case of the graph of an equation.

Highest point on the curve when the tangent is drawn it make angle $0^0$ with the x -axis that point is Maxima.

Lowest point on the curve when the tangent is drawn it make angle $0^0$ with the x -axis that point is Minima.

Calculation:

The function is given as:

  $f\left(x\right)=x^5-4x^3+3x^2-8x-6$ .

The graph of the given equation $f\left(x\right)=x^5-4x^3+3x^2-8x-6$

by collecting the points will be:

Using graphing calculator,

The maxima point is $\left(-1.871,22.741\right)$ .

The minima point is $\left(1.518,-17.162\right)$ .

Hence, the x coordinate of the maxima and minima are Maxima= $-1.871$ and minima point is $1.518$ .