Answer to Question #159567 in Calculus for ghfg

Question

Locate the absolute maximum and minimum for each of the following functions and justify your responses. (Show all work)

1) 𝑓(𝑥) = square root of x^2 − 𝑥 on the interval [−1,1]

2) 𝑔(𝑥) = 𝑥𝑒^2𝑥 on the interval [−2,0]

Answer 1

$Solution: (i)~Given ~ that ~: f(x)=\sqrt {x^2-x } ~on ~[-1,1] \\We~find ~ f'(x)=(2x-1) \frac{1}{2(x^2 -x )^{\frac{1}{2}}} \\Now ~solving ~ for ~ (2x-1) \frac{1}{2(x^2 -x )^{\frac{1}{2}}}=0 \\we ~get~ x=\frac{1}{2} \\ \therefore ~we~ find~ f(x)~ on~~ the ~ points x=-1,\frac{1}{2},1 \\ \therefore we~ get,f(-1)= \sqrt{2}=1.41421 \\~~~~~~~~~~~~~~~~~~~~f(\frac{1}{2})=0.8660~and \\~~~~~~~~~~~~~~~~~~~~~f(0)=0 \\ \therefore (2x-1) \frac{1}{2(x^2 -x )^{\frac{1}{2}}} \\\therefore Absolute~ Maximum~=1.41421~ and ~Absolute ~minimum~=~0 \\(ii)~g(x)=xe^{2x}~on~[-2,0] \\ \therefore g'(x)=2xe^{2x}+e^{2x}=e^{2x}(2x+1) \\ Solving, e^{2x}(2x+1)=0, We~ get, \\x=-\frac{1}{2} \\ \therefore we~find ~ f(x)~at ~points~x=-2,-\frac{1}{2},0 \\ \therefore f(-2)=-0.036631 \\f(-\frac{1}{2})=-\frac{1}{2e}=-0.183939720~and ~ \\f(0)=0 \\\therefore Absolute~ Maximum~=0~ and ~Absolute ~minimum~=~-0.183939720$