Define $$g(x)=\int_{-3}^{3}{f(x-y)f(y)dy}$$, for all real x, here $$f(t)=\left\{ \begin{align} & 1,0\le t\le 1 \\ & 0,elsewhere \\ \end{align} \right\}$$ then A.$$g(x)$$ is not continuous everywhere.B.$$g(x)$$ is continuous everywhere but not differential. C.$$g(x)$$ is continuous everywhere and differentiable everywhere except at $$x=0,1$$. D.$$g(x)$$ is continuous everywhere and differentiable everywhere except at $$x=0,1,2$$.

Question

Define $$g(x)=\int_{-3}^{3}{f(x-y)f(y)dy}$$, for all real x, here  $$f(t)=\left\{ \begin{align}  & 1,0\le t\le 1 \\  & 0,elsewhere \\ \end{align} \right\}$$ then A.$$g(x)$$ is not continuous everywhere.B.$$g(x)$$ is continuous everywhere but not differential. C.$$g(x)$$ is continuous everywhere and differentiable everywhere except at $$x=0,1$$. D.$$g(x)$$ is continuous everywhere and differentiable everywhere except at $$x=0,1,2$$.

Vedantu Content Team · Accepted Answer

Hint: Our first step is to check whether the given function $$g(x)=\int_{-3}^{3}{f(x-y)f(y)dy}$$ is continuous and differential. To check whether the given function $$g(x)=\int_{-3}^{3}{f(x-y)f(y)dy}$$ is continuous and differentiable we can use the mean value theorem. After that if we get discontinuity then we can find the points at which the function is discontinuous. After that we can select the correct option. Complete step-by-step answer:We can write the given function as,$$g(x)=\int_{0}^{1}{f(x-y)dy}$$Now we will make this integration simple by making the substitution given below.$$\begin{align}  & x-y=t \\  & -dy=dt \\ \end{align}$$ After substitution we get,$$g(x)=\int_{x-1}^{x}{f(t)}$$We can see that limits are also changed.$$g(x)=\left\{ \begin{align}  & 0,x\le 0 \\  & x,0 < x < 1 \\  & 2-x,1\le x\le 2 \\  & 0,x > 2 \\ \end{align} \right\}$$Now from the above function definition we can see that the function is continuous and differentiable but at some points. From the function definition it is clear that the function is not differentiable at  $$x=0,1,2$$.Thus, we can conclude that $$g(x)$$ is continuous everywhere and differentiable everywhere except at $$x=0,1,2$$.So, the correct answer is “Option D”.Note: Generally, if a function is differentiable then it is also continuous and a function’s continuity does not guarantee a function’s differentiability. If a function is constant then that function is infinitely differentiable and all its differentials are identical and equal to zero. As an interesting fact zero is a continuous function. Do not forget to change the limits as per the assumption made. Carefully check the points where function is not differentiable.