Question

Complete parts a through c for the given function.

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.