Question

Complete parts a through c for the given function.

f(x) = 4x^5 + 10x^4 - 100x^3 - 4 on [-6,4]

a. Locate the critical point(s) off. Select the correct choice below and, if necessary, fill in the answer box to complete your choice

A. The critical point(s) is(are) at x =

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

B. The function does not have a critical value.
b. Use the First Derivative Test to locate the local maximum/maxima and minimum/minima values. Select the correct choice below and, if necessary, fill in the answer box to complete your choice

A. The local minimum/minima is/are at x = (Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

B. The local maximum/maxima is/are at x = (Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

C. The local maximum/maxima is/are at x = Q and the local minimum/minima is/are at x = (Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

D. There is no local minimum and there is no local maximum. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

c. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

A. The absolute maximum is at x = but there is no absolute minimum. (Use a comma to separate answers as needed. Type integers or simplified fractions.)

B. The absolute maximum is at x = and the absolute minimum is at x = (Use a comma to separate answers as needed. Type integers or simplified fractions.)

C. The absolute minimum is at x = but there is no absolute maximum. (Use a comma to separate answers as needed. Type integers or simplified fractions.)

D.The function has no absolute extrema.

Answer 1

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$   is a maximum and $x=3$ is a minimum

c. The absolute minimum is at   $x = 3$  and the absolute maximum is at  $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out    $20x^2$  and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have   $20x^2 = 0$  , so $x = 0$.

And you would also have  $x^2 +2x-15 = 0$.

You can factor that equation as    $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore   $x=-5 , x=3$.

So the critcal points are at

$x=0,-5,3$

b.  

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0$

Then, $x = -5$   is a maximum and $x=3$ is a minimum

c.

The absolute minimum is at   $x = 3$  and the absolute maximum is at  $x = -5.$

Answer 2

Answer:

(a) THE CRITICAL POINTS ARE (-5, 0, 3) on [-6, 4]

(b) THE LOCAL MINIMUM ARE AT

(-5, 3, 0), THERE ARE NO LOCAL MAXIMUM.

(c) THE ABSOLUTE MINIMUM IS -5

THE ABSOLUTE MAXIMUM IS 0.

(d) THE FUNCTION HAS NO EXTREMA.

Step-by-step explanation:

Given f(x) = 4x^5 + 10x^4 - 100x^3 - 4

We are required to find

(a) THE CRITICAL POINTS

The critical points are are the points where the first derivative vanishes.

That is the values x, where

f'(x) = 0

To find these points, let us differentiate f'(x) with respect to x

f'(x) = 20x^4 + 40x³ - 300x²

The points where f'(x) = 0 are the points where

20x^4 + 40x³ - 300x² = 0

20x²(x² + 2x - 15) = 0

x² = 0 => x = 0 is a point

The other points are

x² + 2x - 15 = 0

(x - 3)(x + 5) = 0

x = 3 and x = -5 are the remaining points.

THE CRITICAL POINTS ARE (-5, 0, 3) which are actually on [-6, 4]

(b) LOCAL MAXIMUM AND LOCAL MINIMUM

The Local Maximum is the critical point where the function is greater than zero, and the Local Minimum is the critical point where the function is less than zero.

f(x) = 4x^5 + 10x^4 - 100x^3 - 4

f(-5) = 4(-3125) + 10(625) + 100(-125)

= -15625 + 6250 - 12500 - 4

= -21879

f(0) = -4 < 0

f(3) = 4(243) + 10(81) - 100(27) - 4

= 972 + 810 - 2700 - 4

= -922 < 0

THE LOCAL MINIMUM ARE AT

(-5, 3, 0), THERE ARE NO LOCAL MAXIMUM.

(c) THE ABSOLUTE MINIMUM IS -5

THE ABSOLUTE MAXIMUM IS 0.