given the function f(x)=-2x^4-16x^3-36x^2, determine all intervals on which f is both decreasing and concave down
1 answer:
Answer:
(0, ∞)
Step-by-step explanation:
The only maximum, local or otherwise, is at x=0. Since the leading coefficient is negative, the function is decreasing and concave down for x-values greater than 0.
0 < x < ∞ . . . . . decreasing and concave down
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Since the leading coefficient of this even-degree function is negative, it will have a generally ∩ shape. It will be decreasing to the right of any local maximum.
The first derivative is negative only for x > 0, so that is the only region where the function is decreasing. Since that x=0 is a local maximum, the function must be concave downward in that neighborhood.
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The attached graph shows the function and its first and second derivatives. The function is concave down where the second derivative is negative, to the right of the right-most point of inflection.
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<em>Additional comment</em>
A graphing calculator provides a quick answer to a question like this. You can find the same answer algebraically.
The first derivative is ...
f'(x) = -8x^3 -48x^2 -72x = -8x(x^2 +6x +9) = -8x(x+3)^2
This has a zero at x=-3 where the graph of it touches the x-axis, but does not change sign. That means f(x) has a flat spot at x=-3, but is not decreasing except to the right of x=0.
The second derivative is ...
f''(x) = -24x^2 -96x -72 = -24(x^2 +4x +3) = -24(x +1)(x +3)
The second derivative will be negative for x < -3 and x > -1, to the left and right of the points of inflection. This means f(x) is concave upward only for -3 < x < -1, and concave downward everywhere else.
The function is concave downward and decreasing for x > 0.
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