Question

Use the alternative form of the derivative to find the derivative at x = c (if it exists). (If the derivative does not exist at c, enter UNDEFINED.) h(x) = |x + 8|, c = -8

Answer 1

Answer:

The derivative of the function does not exist.

Step-by-step explanation:

The alternative form of a derivative is given by:

$f'(c)= \lim_{x \to c} \dfrac{f(x)-f(c)}{x-c}$

Our function is defined as:

h(x)=|x+8|

i.e.  h(x)=   -(x+8) when x+8<0

           and   x+8  when x+8≥0

i.e. h(x)=  -x-8   when x<-8

      and   x+8  when x≥-8

Hence now we find the derivative of the function at c=-8

i.e. we need to find the Left hand derivative (L.H.D.) and Right hand derivative (R.H.D) of the function.

The L.H.D at a point 'a' is calculated as:

$\lim_{x \to a^-} \dfrac{f(x)-f(a)}{x-a}\\\\=\lim_{h \to0} \dfrac{f(a-h)-f(a)}{a-h-a}= \lim_{h\to 0}  \dfrac{f(a-h)-f(a)}{-h}$

Similarly R.H.D is given by:

$\lim_{x \to a^+} \dfrac{f(x)-f(a)}{x+a}\\\\=\lim_{h \to 0} \dfrac{f(a+h)-f(a)}{a+h-a}= \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}$

Now for L.H.D we have to use the function h(x) =-x-8

and for R.H.D. we have to use the function h(x)=x+8

L.H.D.

we have a=-8

$\lim_{x \to (-8)^-} \dfrac{h(x)-h(-8)}{x-(-8)}\\\\= \lim_{h \to0} \dfrac{h(-8-h)-h(-8)}{-8-h-(-8)}= \lim_{h\to 0} \dfrac{h(-8-h)-h(-8)}{-h}$

= $\lim_{h \to 0} \dfrac{8+h-8-0}{-h}= \lim_{h \to 0}\dfrac{h}{-h}=-1$

similarly for R.H.D.

$\lim_{x \to (-8)^+} \dfrac{h(x)-h(-8)}{x-(-8)}\\\\=\lim_{h \to 0} \dfrac{h(-8+h)-h(-8)}{-8+h-(-8)}= \lim_{h\to 0} \dfrac{h(-8+h)-h(-8)}{h}$

$\lim_{h \to 0} \dfrac{-8+h+8-0}{h}=\lim_{h \to 0}\dfrac{h}{h}=1$

Now as L.H.D≠R.H.D.

Hence, the function is not differentiable.