Is Function f(x) Differentiable at x = 4?

[ziddy83]

hey guys, what's up. Its my first time posting so yeah...

Ok I'm attempting to do the following problem...

let f(x) = x^3 for x is less than or eqaul to 4.
(6x^2)-8x when x is greater than 4

i need to see if f(x) is differentiable at x = 4. I tried it through the definition of the derivative as a function. For the first part I am using x^3 and plugging it into f(x) in the def. of the deriv. and i can't seem to figure out what to do with the h I have in the exponent and in the denomenator. i fot as far as...

x^3 lim (a^h-1)/h...I know what x is, 4...but what do i do with the a and the h? please someone help, thanks.

 

[vsage]

Not sure how you would accomplish what you're doing easily with your problem but I would personally see if the derivative at x = 4 was the same at both functions. Note the function is differentiable at a point x if the limit of the function from the positive side is the same as the limit of the function from the negative side.

 

[wais]

Hi there! Welcome to the community and thanks for posting your question.

To determine if a function is differentiable at a specific point, we need to check if the left and right hand derivatives at that point are equal. In this case, we need to check if the left and right hand derivatives at x = 4 are equal.

Let's start with the left hand derivative. Using the definition of the derivative, we have:

f'(4-) = lim (h->0) (f(4-h) - f(4)) / h

= lim (h->0) ((4-h)^3 - 4^3) / h

= lim (h->0) (64 - 12h^2 + h^3 - 64) / h

= lim (h->0) (-12h^2 + h^3) / h

= lim (h->0) (-12h + h^2)

= 0

Now, let's calculate the right hand derivative:

f'(4+) = lim (h->0) (f(4+h) - f(4)) / h

= lim (h->0) ((4+h)^3 - 4^3) / h

= lim (h->0) (64 + 12h^2 + h^3 - 64) / h

= lim (h->0) (12h^2 + h^3) / h

= lim (h->0) (12h + h^2)

= 12

Since f'(4-) = 0 and f'(4+) = 12, the left and right hand derivatives are not equal, which means that the function is not differentiable at x = 4.

I hope this helps! Let me know if you have any other questions. Good luck with your problem!