Find the Critical Number's of the function:?

Question

f(x)= 3√x^2 - x

Answer 1

u = x^2 - x

du/dx = 2x - 1

y = 3u^(1/2)

dy/du = 3/2 u (-1/2) = 3/(2 sqrt u)


dy/dx = dy/du du/dx = 3/(2 sqrtu)(2x-1)

= 3/(2 sqrt (x^2-x)) (2x-1)

= 3(2x-1)/(2 sqrt(x^2-x))

This equals zero if 2x-1 = 0

Find that value for x and plug it in for f(x)

If it doesn't have a value there then the function has no critical value there

My textbook defines a critical point as a point which is included in the domain of f(x) at which the first derivative is zero OR does not exist. If this is correct then places where the denominator equal zero would NOT be critical points, since they would not be in the domain, contrary to what other answers here are saying. Time for me to do some research on whether critical NUMBERS are the same as critical POINTS.

Thanks to puggy below for straightening me out. The function f(x) contains 1 and 0 in its domain; it's f'(x) that doesn't have them in its domain. However, 1/2 is NOT in the domain of f(x) and therefore will NOT give a critical number. This answers site is continually educational to me.

If you didn't mean the whole x^2-x was under the radical, ignore this whole post.

Answer 2

Do you mean the square root of 3 or the square root of x^2? I'm going to assume you mean the square root of 3, since the problem becomes ridiculous if you meant to say the square root of x^2.

f(x) = sqrt(3) x^2 - x

Critical numbers are defined to be where f'(x) = 0 OR where f'(x) is undefined. Keep this in mind next time you want to solve for critical numbers.

Your first step is to find the derivative.

f'(x) = 2 sqrt(3) x - 1

Now, make it 0.

0 = 2 sqrt(3) x - 1
1 = 2 sqrt(3) x
1/(2 sqrt(3)) = x

Therefore, x = 1/(2 sqrt(3)) is the critical number.

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If on the other hand, you meant to say
f(x) = 3 sqrt(x^2 - x), then you'd solve it the same way.

f'(x) = 3(1/2)(x^2 - x)^(-1/2) (2x - 1)
f'(x) = [(3/2)(2x-1)] / [sqrt(x^2 - x)]

Make f'(x) = 0.

0 = [(3/2)(2x-1)] / [sqrt(x^2 - x)]

This is equal to 0 when the numerator is equal to 0.
This is undefined when the denominator is equal to 0.
Critical numbers are defined to be when f'(x) is 0 or when f'(x) is undefined. Let's keep this in mind when continuing.

For the numerator to be 0, (3/2)(2x - 1) = 0
This works out to 2x - 1 = 0, 2x = 1, or x = 1/2

For the denominator to be 0, let
sqrt(x^2 - x) = 0. Square both sides, to get
x^2 - x = 0. Factor,
x(x-1) = 0
Therefore, x = 0 and x = 1.

So your critical numbers are: x = 1/2, 0, and 1

Edit: Looks like most people took the second interpretation. Please be careful when posting questions, as the form of it is extremely important.

Answer 3

f'(x)=3*[1/2(x^2-x)^-1/2*](2x-1)
setting this to zero 2x-1=0 or x=1/2
x(x-1)=0
x=0 or 1
so the critical numbers are 1/2,0 and 1