How do you find concavity, inflection points, and min/max points for the function: #f(x)=x(x^2+1)# defined on the interval [–5,4]?

Question

Christopher Agresta · Accepted Answer

Take the first and second derivative and their zeros.

First, look at the graph so you can get an idea of what you are working with. graph{x(x^2+1) [-10, 10, -5, 5]} At first glance, there don'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)=x^3 + x# #f'(x) = 3x^2 + 1# #x^2 = -1/3# Since #x^2# 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. #f''(x) = 6x# This 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<0# so its concave down and positive for #x>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<0# concave up: #x>0#

Emilia Aguilar · Answer

undefined To find the concavity, inflection points, and min/max points for the function $f(x)=x(x^2+1)$ on the interval $[-5,4]$:

1. **Concavity**: Find the second derivative of $f(x)$, $f''(x)$. If $f''(x) > 0$, the function is concave up. If $f''(x) < 0$, the function is concave down.

2. **Inflection Points**: Inflection points occur where the concavity changes. Set $f''(x) = 0$ and solve for $x$. These are potential inflection points. Test the concavity on either side of these points to confirm if they are inflection points.

3. **Min/Max Points**: To find the min/max points, first find the critical points. These occur where the first derivative, $f'(x)$, is zero or undefined. Then, test the value of the function at these critical points and at the endpoints of the interval to determine if they correspond to minima, maxima, or neither.

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