# To explain why the local maxima and minima of f and g occur at the same value of x .

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 2.3, Problem 58E

a.

To determine

## To explain why the local maxima and minima of f and g occur at the same value of x .

Expert Solution

The local maxima and minima of f and g occur at the same value of x because it is impossible for g to have local maxima and minima at x=a if f do not have local maxima and minima at the same point x=a .

### Explanation of Solution

Given information:

the function is g(x)=f(x)

The local maxima of g will occur at x=a if and only if g(a)g(x) since, g(x)=f(x) so, conversely, g(a)g(x) will be true if and only if f(a)f(x) .

Similarly,

The local minima of g will occur at x=a if and only if g(a)g(x) since, g(x)=f(x) so, conversely, g(a)g(x) will be true if and only if f(a)f(x) .

Hence,

The local maxima and minima of f and g occur at the same value of x because it is impossible for g to have local maxima and minima at x=a if f do not have local maxima and minima at the same point x=a .

b.

To determine

### To express g as a function of x .

Expert Solution

g as a function of x is g(x)=x4+x26x+9

### Explanation of Solution

Given information:

g(x) is the distance between the point (3,0) and the point (x,x2) on the graph of the parabola y=x2 .

Formula used:

Distance between two point (x1,y1) and (x2,y2) is defined by,

d=(x2x1)2+(y2y1)2

Calculation:

g(x) is the distance between the point (3,0) and the point (x,x2) on the graph of the parabola y=x2 .

Therefore,

Use distance formula,

d=(x2x1)2+(y2y1)2g(x)=(x3)2+(x20)2g(x)=x26x+9+x4g(x)=x4+x26x+9

Hence,

g as a function of x is g(x)=x4+x26x+9

c.

To determine

### To find the minimum value of the function g .

Expert Solution

The minimum value of the function g is 2.236 at x=1 .

### Explanation of Solution

Given information:

g(x) is the distance between the point (3,0) and the point (x,x2) on the graph of the parabola y=x2 .

Calculation:

The function is g(x)=x4+x26x+9 from part (b).

Use the principle describe in part (a) that the local maxima and minima of f and g occur at the same value of x .

Compare g(x)=x4+x26x+9 with g(x)=f(x) ,

f(x)=x4+x26x+9

The graph of f is shown below:

There appears to be one local minimum at x=1 .

Hence,

Using maximum command, the local minimum value of y is 5 and this value occurs when x is 1.

Since, the local maxima and minima of f and g occur at the same value of x .

Hence, localminima of g occur at the same value of x that is x=1 .

So, the minimum value of g is obtained as:

g(1)=f(1)g(1)=5g(1)=2.236

Hence,

The minimum value of the function g is 2.236 at x=1 .

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