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Chapter 4: Problem 58

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$f(x)=-x^{4}-2 x^{3}+12 x^{2}$$

Short Answer

Expert verified
Question: Determine the intervals of concavity and inflection points for the function $$f(x) = -x^4 - 2x^3 + 12x^2$$. Answer: The function is concave up in the interval (-2, 1) and concave down in the intervals (-鈭�, -2) and (1, 鈭�). The inflection points are at x = -2 and x = 1.

Step by step solution

01

Find the first derivative of the function

Differentiate the given function with respect to x. $$f'(x) = -4x^3 - 6x^2 + 24x$$
02

Find the second derivative of the function

Differentiate the first derivative with respect to x to find the second derivative of the function. $$f''(x) = -12x^2 - 12x + 24$$
03

Find critical points and inflection points

Find the values of x for which the second derivative is zero or does not exist (critical points) and analyze their intervals to find the concavity. $$f''(x) = -12x^2 - 12x + 24 = 0$$ To solve the quadratic equation, we can factor the equation: $$-12(x^2 + x - 2) = 0$$ $$-12(x - 1)(x + 2) = 0$$ The critical points are x = 1 and x = -2.
04

Test intervals and determine concavity

Test intervals determined by the critical points to see if the second derivative is positive or negative in those intervals: - Test x = -3 (in the interval \((-\infty, -2)\)): $$f''(-3) = -12(-3)^2 - 12(-3) + 24 < 0$$ The second derivative is negative, so the function is concave down in the interval \((-\infty, -2)\). - Test x = 0 (in the interval \((-2, 1)\)): $$f''(0) = -12(0)^2 - 12(0) + 24 > 0$$ The second derivative is positive, so the function is concave up in the interval \((-2, 1)\). - Test x = 2 (in the interval \((1, \infty)\)): $$f''(2) = -12(2)^2 - 12(2) + 24 < 0$$ The second derivative is negative, so the function is concave down in the interval \((1, \infty)\).
05

Write the final answer

Based on our analysis, the function is: - Concave up in the interval \((-2, 1)\) - Concave down in the intervals \((-\infty, -2)\) and \((1, \infty)\) The inflection points, where the concavity changes, are at x = -2 and x = 1.

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