Question
Question: Let \(f:R\to R\) be a differentiable function at \(c\in R\) and f( c ) = 0. If g(x) = If(x)I, then a...
Let f:R→R be a differentiable function at c∈R and f( c ) = 0. If g(x) = If(x)I, then at x = c, g is
(a) differentiable if f’(c) = 0
(b) not differentiable
(c) differentiable if f′(c)=0
(d) not differentiable if f’(c) = 0
Solution
To solve this question, we will first use the definition of differentiability, to find the value of derivative of function g at x = c. then, again we will modify the formula of differentiability for h = 0. And then we will use the concept of differentiability of modulus function. Then, we will discard the options and choose the right option.
Complete step-by-step answer:
Now, in question it is given that f:R→R be differentiable function at c∈R and f( c ) = 0,
and g(x) = If(x)I.
then the function g at x = c will be
g′(c)=h→0limhg(c+h)−g(c) , where h is greater than 0, that is h > 0
Here we have g(x) = I f(x) I
So, g′(c)=h→0limh∣f(c+h)∣−∣f(c)∣
As, it is given that f(c) = 0, so ∣f(c)∣=0
Then, g′(c)=h→0limh∣f(c+h)∣−0
g′(c)=h→0limh∣f(c+h)∣
Also, we can write g’(c) as
g′(c)=h→0limhf(c+h)−f(c), where h > 0
g′(c)=h→0limhf(c+h)−f(c)
As, h→0limhf(c+h)−f(c) denotes f’(c)
So, we can say that h→0limhf(c+h)−f(c)=∣f′(c)∣
g′(c)=∣f′(c)∣
Or, f is differentiable at x = c
Now, if f’(c) = 0 then g(x) is differentiable at x = c otherwise, left hand derivative say ( LHD )at x = c and right hand derivative ( RHD ) at x = c is different.
Now, also we know that modulus function ∣x∣ is differentiable everywhere except at x = 0.
So, we can say that g is differentiable at f′(c)=0, as we proved above that g is differentiable and equals to g′(c)=∣f′(c)∣.
So, the correct answer is “Option c”.
Note: Always remember that if we have to differentiate a function f(x) at point x = a, then we can evaluate the differentiation of f(x) by formula f′(a)=h→0limhf(a+h)−f(a). Also, remember that function ∣x∣ is differentiable everywhere except at x = 0. Try to choose the correct option by discarding wrong options first.