How do you find concavity, inflection points, and min/max points for the function: f(x)=x(x2+1) defined on the interval [−5,4]?

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tempur8x43 · Accepted Answer

Answer: Take the first and second derivative and their zeros.At first glance, there don&#039;t appear to be any max or min on the function. The graph looks concave down to the left and up on the right. Just to be sure, lets do the math. We need to take the first derivative, and that will be easier once we multiply the x through.f(x)=x3+xf′(x)=3x2+1x2=−13Since x2 would need to be negative, there are no real zeros. This means the min an max will be the endpoints, x=−5 and x=4. To get inflection points and concavity, we need to take the second derivative.fx)=6xThis function is linear, so our work becomes pretty easy. The inflection point is x=0 because that is where our expression =0. The expression is negative for x&amp;lt;0 so its concave down and positive for x&amp;gt;0 so its concave up. To find your points, plug your x values into f(x) and solve.min: (−5,−130)max: (4, 68)inflection: (0, 0)concave down: x&amp;lt;0concave up: x&amp;gt;0

How do you find concavity, inflection points, and

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