How to prove that the derivative f this function is |x|

[transgalactic]

how to prove that the derivative of this expression


<br /> f(x)=-\frac12x^2,x&lt;0<br />
<br /> f(x)=\frac12x^2,x&gt;=0 <br />
is f'(x)=|x|

i tried
<br /> \lim _{x-&gt;0^-}\frac{f(x)-f(0)}{x}=\lim _{x-&gt;0^-}\frac{-\frac12x^2-0}{x}=\lim _{x-&gt;0^-}-\frac12x=0\\<br />
<br /> \lim _{x-&gt;0^-}\frac{f(x)-f(0)}{x}=\lim _{x-&gt;0^-}\frac{+\frac12x^2-0}{x}=\lim _{x-&gt;0^+}+\frac12x=0<br />
but i get values
it doesn't show that f'(x)=|x|
??

 

[CompuChip]

You calculated f'(0) and found that it is 0. So this doesn't contradict f'(x) = |x|. Now calculate the derivative for all other x as well

 

[transgalactic]

<br /> \lim _{h-&gt;0^-}\frac{f(x+h)-f(x)}{h}=\lim _{h-&gt;0^-}\frac{-\frac12(x+h)^2+\frac12x^2}{h}=\lim _{h-&gt;0^-}\frac{-\frac12(x^2+2xh+h^2)+\frac12x^2}{h}=\lim _{h-&gt;0^-}\frac{-\frac12(2x+h)}{1}=-x\\<br />
<br /> \lim _{h-&gt;0^+}\frac{f(x+h)-f(x)}{h}=\lim _{h-&gt;0^+}\frac{\frac12(x+h)^2-\frac12x^2}{h}=\lim _{h-&gt;0^+}\frac{\frac12(2xh+h^2)}{h}=\lim _{h-&gt;0^+}\frac{+\frac12(2x+h)}{1}=x\\<br />

what to write in order to finish this prove??

 

[CompuChip]

Almost... you didn't mean to write h \to 0^\pm under the limits, because both limits are two-sided. So, the first one is just
<br /> \lim _{h\to0}\frac{f(x+h)-f(x)}{h}=\lim _{h\to0}\frac{-\frac12(x+h)^2+\frac12x^2}{h}=\lim _{h\to0}\frac{-\frac12(x^2+2xh+h^2)+\frac12x^2}{h}=\lim _{h\to0}\frac{-\frac12(2x+h)}{1}=-x\\<br />
You probably meant to say, that the top line is for x < 0 and the bottom line is for x > 0 (because on the top line you are using the part of the definition for x < 0 and on the bottom line you are using the definition of f for x >= 0).

So now you have shown that f' exists everywhere and is equal to
f&#039;(x) = \begin{cases} -x &amp; \text{ if } x &lt; 0 \\ 0 &amp; \text{ if } x = 0 \\ x &amp; \text{ if } x &gt; 0 \end{cases}

Can you conclude now that f'(x) = |x| ?

 

[transgalactic]

can I ?

 

[quantumdude]

You should know. Compare his expression for f&#039;(x) with the definition of |x|.

 

[Redbelly98]

can I ?

Look at robueler's post concerning y(x) = |x|.