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nordsb [41]
3 years ago
11

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
Mathematics
1 answer:
KIM [24]3 years ago
4 0

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.



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