What are the absolute extrema of $f (x) = x^{5} - x^{3} + x^{2} - 7 x \in [0, 7]$ $f(x)= x^5 -x^3+x^2-7x in [0,7]$ ?

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Answer 1 · 2017-07-02T09:59:31

Minimum: $f (x) = - 6.237$ $f(x) = -6.237$ at $x = 1.147$ $x= 1.147$

Maximum: $f (x) = 16464$ $f(x) = 16464$ at $x = 7$ $x = 7$

We're asked to find the global minimum and maximum values for a function in a given range.

To do so, we need to find the critical points of the solution, which can be done by taking the first derivative and solving for $x$ $x$ :

$f'(x) = 5x^4 - 3x^2 + 2x - 7$

$x ~~ 1.147$

which happens to be the only critical point.

To find the global extrema, we need to find the value of $f(x)$ at $x=0$ , $x = 1.147$ , and $x=7$ , according to the given range:

Thus the absolute extrema of this function on the interval $x in [0, 7]$ is

Minimum: $f(x) = -6.237$ at $x = 1.147$

Maximum: $f(x) = 16464$ at $x = 7$

What are the absolute extrema of f(x)= x^5 -x^3+x^2-7x in [0,7]f(x)=x5−x3+x2−7x∈[0,7] f(x)= x^5 -x^3+x^2-7x in [0,7]?