High School

Problem 1:

Consider the function [tex]f(x) = 12x^5 + 60x^4 - 100x^3 + 2[/tex]. For this function, there are four important intervals: [tex](-\infty, A)[/tex], [tex](A, B)[/tex], [tex](B, C)[/tex], and [tex](C, \infty)[/tex], where [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex] are the critical numbers.

1. Find the critical numbers [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex].
2. At each critical number [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex], determine whether [tex]f(x)[/tex] has a local minimum, a local maximum, or neither.

Type in your answer as LMIN, LMAX, or NEITHER.

Answer :

To find the critical numbers of the function f(x) = 12x^5 + 60x^4 - 100x^3 + 2, we need to find the values of x where the derivative of f(x) is equal to zero or undefined.

Let's start by finding the derivative of f(x):

f'(x) = 60x^4 + 240x^3 - 300x^2

Setting f'(x) equal to zero:

60x^4 + 240x^3 - 300x^2 = 0

Factoring out common terms:

60x^2(x^2 + 4x - 5) = 0

Setting each factor equal to zero:

60x^2 = 0 (gives x = 0)

x^2 + 4x - 5 = 0 (gives two solutions using quadratic formula)

Solving the quadratic equation, we have:

x = (-4 ± √(4^2 - 4(-5))) / 2

x = (-4 ± √(16 + 20)) / 2

x = (-4 ± √36) / 2

x = (-4 ± 6) / 2

The solutions for x are:

x = -5

x = 1

So, the critical numbers are A = -5, B = 0, and C = 1.

Now, to determine the behavior of f(x) at each critical number, we can examine the sign of the derivative f'(x) in the intervals surrounding these critical numbers.

Interval (-∞, A):

For x < -5, the derivative f'(x) = 60x^4 + 240x^3 - 300x^2 > 0. Therefore, f(x) is increasing in this interval.

Interval (A, B):

For -5 < x < 0, the derivative f'(x) = 60x^4 + 240x^3 - 300x^2 < 0. Therefore, f(x) is decreasing in this interval.

Interval (B, C):

For 0 < x < 1, the derivative f'(x) = 60x^4 + 240x^3 - 300x^2 > 0. Therefore, f(x) is increasing in this interval.

Interval (C, ∞):

For x > 1, the derivative f'(x) = 60x^4 + 240x^3 - 300x^2 > 0. Therefore, f(x) is increasing in this interval.

Now, let's determine whether f(x) has a local min, local max, or neither at each critical number.

At A = -5, since f(x) is increasing to the left of A and decreasing to the right of A, f(x) has a local maximum at x = -5.

At B = 0, since f(x) is decreasing to the left of B and increasing to the right of B, f(x) has a local minimum at x = 0.

At C = 1, since f(x) is increasing to the left of C and increasing to the right of C, f(x) does not have a local min or local max at x = 1.

Therefore, the answers are:

A = -5 corresponds to a local maximum (UMAX).

B = 0 corresponds to a local minimum (LMIN).

C = 1 corresponds to neither a local min nor local max (NETHEA).

Learn more about critical numbers here :

https://brainly.com/question/31339061

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