Question

Determine the absolute maximum value of f(x)= xsqrt(1-x^2)-3

Answer 1

on [-1,1]

Answer 2

\[f(x)=x \sqrt{1-x^2}-3\] first we note the function is continuous on [-1,1]. Then we find the critical numbers:
\[f'(x)=\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}\]The critical numbers are the values of the function where the derivative equals zero. Thus,\[2x^2=1\]And,\[x=\pm1/\sqrt{2}\]Now we check the value of the function at the end points x=-1 and x=1:\[f(1)=-3; f(-1)=-3\]The largest of -3,1/sqrt(2) and -1/sqrt(2) is your global maximum. The absolute max is \[\frac{+1}{\sqrt{2}}\]

Answer 3

whoops. the value of the absolute max occures at 1/sqrt2. So plug in x=1/sqrt2 into f(x) to get the max value. I get f(x)=-5/2

Answer 4

omggg hahaha i got up all the way to 1/sqrt2 i didnt know you had to do that last step

Answer 5

lol yeah I almost forgot too. Make sure you evaluate -3 and -1/sqrt2 as well to see which is biggest/smallest. I skipped that step above too, but the solution is correct. f(1/sqrt2)=-5/2 is the max.

Answer 6

Can you elp me with a step on this next problem? i dont wanna now the answer im just getting stuck and need a push

Answer 7

know*

Answer 8

yeah sure

Answer 9

its find all the critical points of the function f(x)= 2sin(x)-|x| on the interval (-pi,pi) ok so i got the derivative and i set it equal to 0 right?

Answer 10

yep

Answer 11

but you have to deal with your absolute value of x term

Answer 12

so, i would split this one into two problems:
g(x)=2sinx-x on [0,pi) and
h(x)=2sinx+x on (-pi,0)

Answer 13

ok... i just took the derivative of the |x| which is x/sqrtx^2.. and when i set it equal to 0, x= pi/3

Answer 14

No. The derivative of abs(x) is wrong.

Answer 15

ohh it is? :/ damn

Answer 16

If \[x \ge0\]Then\[\left| x \right|=x\]

Answer 17

If x<0 then\[\left| x \right|=-x\]

Answer 18

So your derivative is either +1 or -1 depending on whether x is greater than (or equal to) zero or not

Answer 19

So we have two functions here, since x takes on both positive and negative values:
g(x)=2sinx-x [0,pi)
h(x)2sinx+x (-pi,0)

Answer 20

so take the derivative of each function ad=nd then set that = to 0 and just combine the two?

Answer 21

Taking the derivatives and setting equal to zero:
On the interval [0,pi)
0=2cosx-1
cosx=1/2
x=pi/3
It's in the interval so we're good.

and for the interval (-pi,0)
0=2cosx+1
cosx=-1/2
x=-2pi/3

Answer 22

Your critical numbers should be \[x=\pi/3, -2\pi/3\]

Answer 23

and 0! :D

Answer 24

and 0! :D

Answer 25

x=0 is not a critical value of this one

Answer 26

yes it is

Answer 27

Because the derivatives don't equal zero when x=0

Answer 28

the answer was pi/3, -2pi/3 and 0 :/

Answer 29

was the question asking just for the critical values?

Answer 30

yea. find all critical values on the interval (-pi,pi)

Answer 31

Oh crap! I know why

Answer 32

A critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0. The function actually isn't differentiable at x=0 because the absolute value function isn't differentiable here . You're correct x=0 too! :)

Answer 33

Forgot about that part of the definition

Answer 34

have you seen why abs(x) have no derivative at zero?

Answer 35

yes :)!

Answer 36

cool :))

Answer 37

thanks for reminding me how to do this stuff lol

Answer 38

thanks for helping me with my steps! and teaching me more than my professor:D