Help me look for maximums/minimums and inflections points as well as how to graph this?

Question

I would appreciate it if you could help me solve for whatever the local maximum and minimums there are as well for any absolute values as well as the inflection points.

Answer 1

since the highest power term has positive coefficient and is even

when x is large negative or x is large positive f(x) is positive large.

notice there are f(x)=0 points at x= -√2 and x=√2

draw up and to left line from (-√2,0) indicating line continues up and left as x decreases

draw up and to right line from (√2,0) indicating line continues up and right as x increases.

notice there is an f(x)=0 point at x=0.

since f(x)<0 for x=(-√2)⁺, then from (-√2,0) to (0,0),, f(x) goes below x-axis and comes back up to (0,0)

since f(x)<0 for x=(√2)^-, then from (√2,0) back to (0,0) f(x) goes below x- axis and comes back up to (0,0)

derivative is f'(x)=4x³-4x,,,derivative has zeroes at -1,0,1 the two points where f(x) has minimum in the rough description of the graph are known= -1 and +1. The values are at x=-1 ,,,f(x)= -1 and at x=1,,,f(x)=-1

the minimum points are (-1,-1) and (1,-1), local minimums both,.and both equaling absolute minimum for f(x)

In the rough description of graph, the point (0,0) would be a local maximum and it is, it is not absolute maximum as f(x) has larger values then f(x)=0

To verify minimums and maximums and to check for inflections, second derivative f''(x)=12x²-4.

At x=-1 and x=1., f''(x)>0 so curve is concave up (and f"(x)=0) means minimum. At x=0 f''(x)<0 so curve is concave down (and f'(x)=0) means maximum. f(x) has inflection (going from concave up to concave down and vice versa) when f''(x)=0. Here f''(x)=0 at x= -(1/√3√) and at x=(1/√3)

Answer 2

Max and mins come from taking the first derivative, setting it equal to zero and solving. If the first derivative goes from negative to positive it is a min. Positive to negative is a max.

Inflection points come from taking the second derivative and setting it equal to zero and solving.

For taking the first derivative, it is better to leave it in the original form, not factored. In the factored form, you would have to use the product rule which is more complex.