Question

f(x)=x^2+2x-1

Prove that it's differentiable in [0,1]

Using: Lim h-->0+ (f(a+h)-f(a))/h = Lim h--> 0- (f(a+h)-f(a))/h

Answer 1

Prove that there exists a derivative and that its equivalent on both sides, throughout that interval.

Answer 2

*it's

Answer 3

How would you prove the left and right hand derivatives are equivalent on both sides throughout the interval specifically?

Answer 4

Note that:
\[
\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=
\lim_{h\to 0^+}\frac{(x+h)^2+2(x+h)-1-(2x^2+2x-1)}{h}
\]Show that this is equal to:
\[
\lim_{h\to 0^+}\frac{(x+h)^2+2(x+h)-1-(2x^2+2x-1)}{h}=2x+2
\]And, that:
\[
\lim_{h\to 0^-}\frac{(x+h)^2+2(x+h)-1-(2x^2+2x-1)}{h}=2x+2
\]And then you have proved it for all real numbers. Clearly \([0,1]\subset\mathbb{R}\). So this holds.

Answer 5

Ok I get it! Because h--> 0 from both sides is just 0 since it's so close it's negligible?

Answer 6

By the way, is y=x^(2/3) a polynomial?

Answer 7

Essentially, but it's not always true of all functions.

And, no, it is not.