ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: pHysiX on May 03, 2009, 08:07:29 pm

Title: some random calculus
Post by: pHysiX on May 03, 2009, 08:07:29 pm: Consider the function $\frac{x}{{1 - |x|}}$

1) Give a reason why the quotient rule for differentiation cannot be used to find g'(0)
2) Determine whether or not g is differentiable at x=0 by evaluating separately the left and right hand limits:
$\mathop {\lim }\limits_{h \to {0^ - }} \frac{{g(h) - g(0)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \frac{{g(h) - g(0)}}{h}<br />$
--------------------------------------------------------------
3) Why is it that derivatives do not exist at end points?
*My reasoning is that the limits at the end points do not exist*

Thanks heaps in advance =]
Title: Re: some random calculus
Post by: TrueTears on May 03, 2009, 09:02:52 pm: 2) g(x) is differentiable at x = 0

Consider the original function $g(x) = \frac{x}{1-|x|}$

when $x > 0$ $g'(x) = \frac{1}{(x-1)^2}$

and when $x < 0$ $g'(x) = \frac{1}{(x+1)^2}$

Sketching these 2 truncas' with the respective domains shows that at x = 0 the graph is continuous.

Now consider the left and right limits

When x is approaching 0 from the negative we are dealing with $\frac{1}{(x+1)^2}$
$lim_{x \to 0^-} \frac{1}{(x+1)^2} = 1$

When x is approaching 0 from the positive we are dealing with $\frac{1}{(x-1)^2}$

$lim_{x \to 0^+} \frac{1}{(x-1)^2} = 1$

$\therefore g(x)$ is differentiable at x = 0
Title: Re: some random calculus
Post by: Mao on May 03, 2009, 10:38:47 pm: 1. |x| is not differentiable at x=0, hence the quotient rule cannot be employed here.

2. as above shown by TT (however, to be pedantic, you should also show g to be continuous at x=0)

3. cannot draw a tangent through an endpoint (one of the left hand or right hand limit does not exist, hence not differentiable)