Question

Use the figure below to estimate the indicated derivatives, or state that they do not exist. If a derivative does not exist, enter dne in the answer blank. The graph of f(x) is black and has a sharp corner at x = 2. The graph of g(x) is blue.

Answer 1

F(x)=3x/2 for 0≤x≤2 
.....=6 - 3x/2 for 2<x≤4 
g(x) = -x/4 + 1 for 0≤x≤4 and g'(x)=-1/4 
so h(x)= f(g(x)) = (3/2)(-¼x+1)=-3x/8 + 3/2 for 0≤x≤2 
for x=1, h'(x)=-3/8 so h'(1)=-3/8 

When x=2, g(2)=1/2 so h'(2)=g'(2)f '(1/2)= -(1/4)(3/2)=-3/8 

When x=3, h(x)=6 - (3/2)(1 - x/4) = 9/2 +3x/8 
h'(x)=3/8 so h'(3) = 3/8

Answer 2

Answer:

f'(x) doesn't exist at x=2 in (0,4)

g'(x) exist over (0,4). $g'(x)=-\dfrac{3}{4}$

Step-by-step explanation:

We are given a graph and need to find the point derivative does not exist.

Black: Graph of f(x)

Blue: Graph of g(x)

In interval (0,4)

f(x) is graph of absolute function. At x=2, f(x) has vertex or corner.

As we know function is not differential at corner.

Hence, f'(x) doesn't exist at x=2 in (0,4)

g(x) is graph of straight line. Derivative of straight line is constant. g(x) is differential over (0,4)

Hence, g'(x) exist over (0,4). $g'(x)=-\dfrac{3}{4}$