Critcal point and extrema

[Roadrunner2015]

Equation is:
f(x) = x(2x-1)^1/3 on [-1,1]

I know the first step to get the critical points is to find f'(x) but the fractional power is throwing me off. My first guess is to make the equation into a cube root of x(2x-1), but to find the derivative from that I am lost. Could somebody please help me through this, thank you very much!

 

[stapel]

Equation is: f(x) = x(2x-1)^1/3 on [-1,1]

I know the first step to get the critical points is to find f'(x) but the fractional power is throwing me off.

Ouch! They were supposed to have taught you about exponents back in algebra!

To learn about fractional (that is, rational) powers, try here. With respect to derivatives, fractional powers obey the same rules as whole-number powers: d(x^n)/dx = n*(x^(n-1)). It doesn't matter if n is a whole number or a fraction.

 

[Roadrunner2015]

Ouch! They were supposed to have taught you about exponents back in algebra!

To learn about fractional (that is, rational) powers, try here. With respect to derivatives, fractional powers obey the same rules as whole-number powers: d(x^n)/dx = n*(x^(n-1)). It doesn't matter if n is a whole number or a fraction.

I'm on a math forum for help and I get linked to google. That's fun.

 

[HallsofIvy]

Usually, one of the first things you learn in Calculus is that the derivative of \(\displaystyle x^n\) is \(\displaystyle nx^{n-1}\). Because of that, most people, if asked to differentiate \(\displaystyle \sqrt[3]{x}\) would write it as \(\displaystyle x^{\frac{1}{3}}\), not the other way around.

 

[Deleted member 4993]

I'm on a math forum for help and I get linked to google. That's fun.

That's because we are NOT here to teach you ... we are here to coach you through your homework. You have to read and learn - hence those sites in Google for reference.

It is fun to force some body to work - when they don't want to do the work!!

 

[Quaid]

f(x) = x(2x-1)^1/3 on [-1,1]

My first guess is to make the equation into a cube root of x(2x-1)

Hello Roadrunner:

Your guess above is a mistake because you're not given the cube root of x(2x-1). Don't change function f. (If you do, you're no longer working the same exercise.)

The cube root of (2x-1) is being multiplied by x. In other words, function f is defined by the product of x times (2x-1)^(1/3).

To take the derivative of a product, you need to use the Product Rule. Have you used the Product Rule, yet?

Additionally, the factor (2x-1)^(1/3) is a composition of a power function with a linear function (i.e., 2x-1 is the inner function, and raising to the 1/3rd power is the outer function). To take the derivative of a composite function, you need to use the Chain Rule. Are you familiar with it?

The best way to get good help quickly, at this tutoring site, is to follow the posting guidelines. Please take time to read the summary page; here's a link.

Cheers :cool: