find-the-absolute-maximum-and-minimum-values-of-the-function-h-on-the-interval-1

Question

Find the absolute maximum and minimum values of the function hh on the interval ﻿[−1,3][-1, 3][−1,3]﻿.﻿h(x)=5x3e−2x h(x) = $$5x^3$$ $$e^{-2x} h(x)=5x3e$$−2x﻿

Accepted Answer

Absolute maximum=1358e−3 at x=32,Absolute minimum=−5e2 at x=−1 	ext{Absolute maximum}=\frac{135}{8}$$e^{-3}$$	ext{ at }x=\frac32,\quad	ext{Absolute minimum}=$$-5e^2$$	ext{ at }x=-1

Answer

Absolute maximum=−1358e−3 at x=−1,Absolute minimum=5e2 at x=32	ext{Absolute maximum}=-\frac{135}{8}$$e^{-3}$$	ext{ at }x=-1,\quad	ext{Absolute minimum}=$$5e^2$$	ext{ at }x=\frac32

Answer

Absolute maximum=−1358e−3 at x=32,Absolute minimum=5e2 at x=−1	ext{Absolute maximum}=-\frac{135}{8}$$e^{-3}$$	ext{ at }x=\frac32,\quad	ext{Absolute minimum}=$$5e^2$$	ext{ at }x=-1

Answer

Absolute maximum=1358e−3 at x=−1,Absolute minimum=−5e2 at x=32	ext{Absolute maximum}=\frac{135}{8}$$e^{-3}$$	ext{ at }x=-1,\quad	ext{Absolute minimum}=$$-5e^2$$	ext{ at }x=\frac32

Find the absolute maximum and minimum values of the function hh on the interval [−1,3][-1, 3][−1,3].h(x)=5x3e−2x h(x) = 5x^3 e^{-2x} h(x)=5x3e−2x

Find the absolute maximum and minimum values of the function $h$ on the interval $[-1, 3]$ .
$h(x) = 5x^3 e^{-2x}$