Derivative of a piecewise function

[MatinSAR]

Homework Statement

Finding derivative of function $f$ which is a piecewise function.

Relevant Equations

$f'(a)=\lim_{x \rightarrow a} {\frac {f(x)-f(a)} {x-a}}$

$f(x) = \begin{cases} \frac {x^3-x^2}{x-1} & \text{if } x\neq1 \\ 1 & \text{if } x=1 \end{cases}$
Find $f'(1)$.

Using derivative definition:
$f'(1)=\lim_{x \rightarrow 1} {\frac {\frac {x^3-x^2}{x-1}-f(1)} {x-1}}$

$f'(1)=\lim_{x \rightarrow 1} {\frac {\frac {x^3-x^2}{x-1}-1} {x-1}}$

$f'(1)=\lim_{x \rightarrow 1} {\frac {x^3-x^2-x+1} {(x-1)^2}}$

Apply L'Hôpital's rule two times:
$f'(1)=\lim_{x \rightarrow 1} {\frac {6x-2} {2}}$
$f'(1)=2$

Can someone please tell me if I'm wrong …

Can I say that $f(x) = \begin{cases} \frac {x^3-x^2}{x-1} & \text{if } x\neq1 \\ 1 & \text{if } x=1 \end{cases}$ graph is similar to $y=x^2$ but it's undefined at $x=1$? If it's true then I can find $f'(1)=2x=2$ far faster.

 

[PeroK]

$f(x) = x^2$? True or false?

 

[MatinSAR]

$f(x) = x^2$? True or false?

No these functions have different domains, so they're not equal.

 

[PeroK]

No these functions have different domains, so they're not equal.

The domain is $\mathbb R$ in both cases.

It's true. The function you are given is plain old $x^2$, wearing a thin disguise!

 

[MatinSAR]

The domain is $\mathbb R$ in both cases.

It's true. The function you are given is plain old $x^2$, wearing a thin disguise!

Oh! I've forgotten that we've defined $f(1)=1$. So both ways were correct, Do you agree?

 

[PeroK]

Oh! I've forgotten that we've defined $f(1)=1$. So both ways were correct, Do you agree?

Both methods are valid. But, the first method is unnecessarily complicated.

As an aside, I've never liked the term piecewise function, as there is no such thing, IMO. You can define a function piecewise. But, as in this case, that doesn't mean that you have to define it piecewise. In this case, you could equally well have written $f(x) = x^2 \ (\forall x)$.

I wonder whether whoever set the question realised this.

 

[MatinSAR]

Both methods are valid. But, the first method is unnecessarily complicated.

Yes.

As an aside, I've never liked the term piecewise function, as there is no such thing, IMO. You can define a function piecewise. But, as in this case, that doesn't mean that you have to define it piecewise. In this case, you could equally well have written $f(x) = x^2 \ (\forall x)$.

I wonder whether whoever set the question realised this.

I've understand it. Thank you for your valuable help and time.