What are the absolute extrema of $f (x) = x^{3} - 3 x + 1 \in [0, 3]$ $f(x)=x^3 - 3x + 1 in[0,3]$ ?

Question

What are the absolute extrema of $f (x) = x^{3} - 3 x + 1 \in [0, 3]$ $f(x)=x^3 - 3x + 1 in[0,3]$ ?

1 Answer

Impact of this question

4185 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-25T21:09:26

On $[0, 3]$ $[0,3]$ , the maximum is $19$ $19$ (at $x = 3$ $x=3$ ) and the minimum is $- 1$ $-1$ (at $x = 1$ $x=1$ ).

Explanation:

To find the absolute extrema of a (continuous) function on a closed interval, we know that the extrema must occur at either crtical numers in the interval or at the endpoints of the interval.

$f (x) = x^{3} - 3 x + 1$ $f(x) = x^3-3x+1$ has derivative

$f'(x) = 3x^2-3$ .

$3x^2-3$ is never undefined and $3x^2-3=0$ at $x=+-1$ .

Since $-1$ is not in the interval $[0,3]$ , we discard it.

The only critical number to consider is $1$ .

$f(0) = 1$

$f(1) = -1$ and

$f(3) = 19$ .

So, the maximum is $19$ (at $x=3$ ) and the minimum is $-1$ (at $x=1$ ).

Science

Math

Humanities

... and beyond

What are the absolute extrema of $f (x) = x^{3} - 3 x + 1 \in [0, 3]$ $f(x)=x^3 - 3x + 1 in[0,3]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What are the absolute extrema of f(x)=x^3 - 3x + 1 in[0,3]f(x)=x3−3x+1∈[0,3]f(x)=x^3 - 3x + 1 in[0,3]?

1 Answer

Explanation:

Related questions

Impact of this question

What are the absolute extrema of $f (x) = x^{3} - 3 x + 1 \in [0, 3]$ $f(x)=x^3 - 3x + 1 in[0,3]$ ?