Question

Consider g(x)=square root(x)−x on the interval [0,4]. At what value of x does the maximum and min occur?

Answer 1

\[g(x)=\sqrt{x}-x=x^{1/2}-x\]\[g'(x)=(1/2)x^{-1/2}-1=\frac{1}{2\sqrt{x}}-1\]
max/min occur when g'(x) = 0, so\[\frac{1}{2\sqrt{x}}-1=0\]\[1-2\sqrt{x}=0\]\[2\sqrt{x}=1\]\[4x=1\]\[x=1/4\]

Answer 2

I am just wondering if that is a minimum or maximum

Answer 3

It is a maximum I just graphed it on my calc. But it looks like a log function. I never kneew log functions have a max or min point

Answer 4

I guess it is the place that it switches directions?!?

Answer 5

you can work that out by taking the 2nd derivative. if it is -ve then you have a maxima

Answer 6

Oh thanks

Answer 7

so what about the minimum

Answer 8

im confused

Answer 9

2nd derivative will be +ve for a minimum

Answer 10

2nd derivative = 0 implies an inflection point

Answer 11

the equation you gave has just one maximum that occurs at x=0.25

Answer 12

oooohhh i think i get it .. thnx