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Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

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BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

In Problems 31-36, both a function and its derivative are given. Use them to find critical values, critical points, intervals on which the function is increasing and decreasing, relative maxima, relative minima, and horizontal points of inflection; sketch the graph of each function.

y = x 2 ( x 5 ) 3   dy =  5 x ( x 2 ) ( x 5 ) 2

To determine

To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of function y=x2(x5)327 and the derivative dydx=5x(x2)(x5)227 and also sketch the graph.

Explanation

Given Information:

The provided function is y=x2(x5)327 and the derivative is dydx=5x(x2)(x5)227.

Formula Used:

The simple power rule of derivative,

ddx(xn)=nxn1

To obtain the relative maxima and minima of a function,

1. Find the first derivative of the function.

2. Set the derivative equal to 0 to obtain the critical points.

3. Evaluate f(x) at values of x to the left and one to the right of each critical point to develop a sign diagram.

(a) If f(x)>0 to the left and f(x)<0 to the right of the critical value, the critical point is a relative maximum point.

(b) If f(x)<0 to the left and f(x)>0 to the right of the critical value, the critical point is a relative minimum point.

Calculation:

Consider the provided function y=x2(x5)327,

The critical values are the only values at which the graph can have turning points, the derivative cannot change sign anywhere except at the critical value.

Hence, there will no change in the values of critical values as in the derivative graph.

Consider the provided derivative,

dydx=5x(x2)(x5)227

Put the value of y=0,

y=dydx=5x(x2)(x5)227=0

Evaluate the values of x from the equation:

5x(x2)(x5)227=0x=0,2,5

Hence, the value(s) of x are x=0,2,5.

Evaluate the values of the original functions with the critical values:

Substitute 0 for x in y=x2(x5)327,

y=(0)2(05)327=027=0

Hence, x=(0,0) is a critical point.

Put x=2 in the equation y=x2(x5)327,

y=(2)2(25)327=4(27)27=4

Hence, x=(2,4) is a critical point.

Put x=5 in the equation y=x2(x5)327,

y=(5)2(55)327=25(0)27=0

Hence, x=(5,0) is a critical point

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