Answer:
Derivative \(f'(x)\) does not exist at \(x=2\) in (0,4)
and Derivative \(g'(x)\) exists over (0,4)
Step-by-step explanation:
Given information
\(h(x)=f(x).g(x)\\\)
So, derivative
\(h'(x)=f(x).g'(x)+g(x).f'(x)\)
From the above equation
\(f'(x)=3/2\) at \(0\leq x\leq 2\)
\(f'(x)=0\) at \(x=2\)
\(f'(x)=-3/2\) at \(2\leq x\leq 4\)
From the graph
\(f(x)\) is graph of absolute function at \(x=2\)
Hence, derivative does not exist at \(x=2\) in (0,4)
Now \(g(x)\) is the graph of straight line and derivative of straight line is constant so, \(g(x)\) is differential over (0,4).
\(g'(x)\) exists over (0,4)
\(g'(x)=-3/4\)
Derivative does not exist at \(x=2\) in (0,4)
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