derivatives: approximate h'(3) for h(x) = x^3 + x^1/3

[Cheryl Hong]

I have this problem and I can't figure out what my teacher wants me to do.

h(x) = x^3 + x^1/3

Approximate h'(3) with the |delta x| equaling 0.01. Write an inequality on which h'(3) lies.

I can find the derivative of h(x). it's h'(x)= 3x^2 - (1/3)x^(-2/3)

I'm not sure what to do with the |delta x| to find h'(3)

 

[galactus]

It appears they want you to use the tangent line approximation and the definition of a derivative.

\(\displaystyle \L\\f(x_{0}+{\Delta}x)\approx{f(x_{0})+f'(x_{0}){\Delta}x\)

See what you get?. Look familiar?.

\(\displaystyle \L\\\frac{f(x_{0}+{\Delta}x)-f(x_{0})}{{\Delta}x}\)


\(\displaystyle \L\\f(x_{0}+{\Delta})=(3+.01)^{3}+(3+.01)^{\frac{1}{3}}\)

 

[Cheryl Hong]

derivatives

Yeah
got that part. thanks.
But I'm not sure how to get the inequality from it. I could get a regular line with the rate of change. does it deal with that?

 

[galactus]

What was the answer you got with the differential?. What was the answer with the actual derivative?.

The differential answer is slightly higher than the actual.

Construct the inequality from that.

 

[Cheryl Hong]

derivatives

:idea: thanks