Question

Discuss the continuity and differentiability of the function
f(x) = x ^3/5

Answer 1

A function is said to be continuous, if for every 'c' in the domain of the function 'f', f(c) is defined and has a finite value with limx-->c f(x) = f(c)

Discontinuity occurs when the function experiences gaps. Now let's test the continuity of the given function f(x) = x⅗

Now, limx-->c x⅗ = c⅗ ....1

Also, f(c) = c⅗ ....2

Since, 1 = 2 we conclude that function f(x) is continuous. The graph looks something like this.

Now let's see whethe f(x) is differentiable. For a function to be differentiable at a point 'c' it should follow the condition below.

f'(c) = limh-->c [f(x + h) - f(x)]/h

where f'(c) is the derivative of the function at c

Let's see, f'(x) = ⅗x(-⅖) So,

f'(c)= ⅗c-⅖

now let's find the limit.

limh-->c [((x+h)3/5) - x3/5 ]/h

Substituting for the limits and using binary expansion we get the limit value to be ...

⅗c-⅖

Which is equal to f'(c). Hence the function is differentiable as well. You can check by substituting random values for c and check on your own.

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