How do you find all critical point and determine the min, max and inflection given $f(x)=x^4$ ?

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Answer 1 · 2016-10-26T19:38:21

$f(x)=x^4$

Domain of $f$ is $(-oo,oo)$ .

A critical number for $f$ is a number $c$ in the domain of $f$ at which $f'(c)$ does not exist or $f'(c)=0$

$f'(x) = 4x^3$ is defined for all $x$ and is $0$ at $x=0$ .

The only critical number for $f$ is $0$ .

(If your treatment of calculus says that a critical point is a point on the graph, then the critical point is $(0,0)$ )

Because $f'(x)$ is negative for $x < 0$ and positive for $x > 0$ , we know that $f(0) = 0$ is a local minimum.

$f''(x) = 12x^2$ which is always positive.

Since the sign of $f''$ never changes, the concavity of $f$ never changes, so there are no inflection points.

How do you find all critical point and determine the min, max and inflection given f(x)=x^4f(x)=x^4?