Prove f(x) is differentiable at x=1 where f(x)=2x^2 x=<1, =4x-1 x>1

[sara_87]

prove f(x) is differentiable at x=1:
f(x)=2x^2 x(less than or equal to)1
4x-1 x>1

 

[PhY]

do you mean:
d/dx:= d/dx(2x^2x) * d/dx(2x) ala function of a function?

i'm probably wrong.

 

[sara_87]

no, i mean it's in piecewise form with the big curly brackets

 

[PhY]

You Mean Integration.

Argh, My Integration is Rusty.
its the opposite of Differentiation.

so 2x would be x^2
2x^2x ...i don't know, because its F(X)^G(X).

You need to hear from somebody else on this.

 

[sara_87]

yeah i think i do need to hear from somebody else on this cos u didnt understand the question ;)
i have to prove that it is differentiable at x=1 it has nothing to do with integration.
thanks anyway

 

[cristo]

The derivative of a function at a point can be expressed as the limit of an expression. You should be able to get two limits; one for each branch of the function. If these are the same, then the function is differentiable at x=1.

 

[JasonRox]

It's amazing how complicated one can make a question when it is meant to be simple. (Not meant to the OP.)

 

[sara_87]

The derivative of a function at a point can be expressed as the limit of an expression. You should be able to get two limits; one for each branch of the function. If these are the same, then the function is differentiable at x=1.

thanx v much i think i can do it now.

 

[CompuChip]

You might also want to check that it is continuous. For example,
$$f(x) = \left\{<br /> \begin{array}{rl}<br /> 0 & \text{ if } x \le 0 \\<br /> 1 & \text{ if } x > 0<br /> \end{array}$$
will give you 0 for the derivative when approaching from the left or right to zero, though at x = 0 the function is not continuous at all.

 

[HallsofIvy]

You Mean Integration.

Argh, My Integration is Rusty.
its the opposite of Differentiation.

so 2x would be x^2
2x^2x ...i don't know, because its F(X)^G(X).

You need to hear from somebody else on this.

No, he meant differentiation. Determine whether
$$f(x) = \left\{ \begin{array}{rl} 2x^2 & \text{ if } x \le 1 \\ 4x-1 & \text{ if } x > 1 \end{array}$$
is differentiable at x= 1

Compuchip's suggestion is etremely good here: a function can't be differentiable if it is't continuous at the point! What is the limit of f as you approach 1 from the left? What is the limit as you approach from the right?

 

[Gib Z]

Argh, My Integration is Rusty.
its the opposite of Differentiation.

Or more precisely, $$\frac{d}{dx}\int^x_a f(t) dt = f(x)$$ where a is a constant. "The Opposite of differentiation" is what people told me before I started integral calculus as well, and it screwed up my understanding a heap load.

2x^2x ...i don't know, because its F(X)^G(X).

You need to hear from somebody else on this.

Just in case you want to know, you let y=2x, and express the remaining integral in terms of the exponential and logarithmic functions.

 

[CompuChip]

I still don't see why you would want to do integration.
It's a partwise defined function, all you need to do is show that the function and the derivative are continuous.