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Math  /  Algebra

QuestionNYA Module 6: Problem 1 (1 point)
The function f(x)=−4x3+24x2−36x−2f(x)=-4 x^{3}+24 x^{2}-36 x-2 is increasing on the interval ( 1,3 ). It is decreasing on the interval (−∞(-\infty, □\square ) and the interval ( □\square , ∞\infty ).
The function has a local maximum at □\square
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Studdy Solution

STEP 1

1. The function f(x)=−4x3+24x2−36x−2 f(x) = -4x^3 + 24x^2 - 36x - 2 is a polynomial function.
2. To determine intervals of increase and decrease, we need to find the derivative of the function and analyze its sign.
3. Critical points occur where the derivative is zero or undefined, and these points help determine intervals of increase and decrease.
4. Local maxima and minima occur at critical points where the derivative changes sign.

STEP 2

1. Find the derivative of the function f(x) f(x) .
2. Determine the critical points by setting the derivative equal to zero.
3. Analyze the sign of the derivative to determine intervals of increase and decrease.
4. Identify the local maximum based on the sign change of the derivative.

STEP 3

Find the derivative of the function f(x)=−4x3+24x2−36x−2 f(x) = -4x^3 + 24x^2 - 36x - 2 :
f′(x)=ddx(−4x3+24x2−36x−2) f'(x) = \frac{d}{dx}(-4x^3 + 24x^2 - 36x - 2)
Using standard differentiation rules:
f′(x)=−12x2+48x−36 f'(x) = -12x^2 + 48x - 36

STEP 4

Determine the critical points by setting the derivative equal to zero:
−12x2+48x−36=0 -12x^2 + 48x - 36 = 0
Simplify by dividing the entire equation by −12-12:
x2−4x+3=0 x^2 - 4x + 3 = 0
Factor the quadratic equation:
(x−3)(x−1)=0 (x - 3)(x - 1) = 0
The critical points are:
x=3andx=1 x = 3 \quad \text{and} \quad x = 1

STEP 5

Analyze the sign of the derivative to determine intervals of increase and decrease. Use test points in each interval determined by the critical points x=1 x = 1 and x=3 x = 3 .
1. Interval (−∞,1)(- \infty, 1): Choose x=0 x = 0 f′(0)=−12(0)2+48(0)−36=−36 f'(0) = -12(0)^2 + 48(0) - 36 = -36 (negative, so decreasing)
2. Interval (1,3)(1, 3): Choose x=2 x = 2 f′(2)=−12(2)2+48(2)−36=12 f'(2) = -12(2)^2 + 48(2) - 36 = 12 (positive, so increasing)
3. Interval (3,∞)(3, \infty): Choose x=4 x = 4 f′(4)=−12(4)2+48(4)−36=−36 f'(4) = -12(4)^2 + 48(4) - 36 = -36 (negative, so decreasing)

STEP 6

Identify the local maximum based on the sign change of the derivative. The function changes from increasing to decreasing at x=3 x = 3 , indicating a local maximum.
The intervals of decrease are (−∞,1)(- \infty, 1) and (3,∞)(3, \infty).
The function has a local maximum at x=3 x = 3 .
The intervals of decrease are (−∞,1)(- \infty, 1) and (3,∞)(3, \infty). The function has a local maximum at x=3 x = 3 .

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