Question

Find the values of m and b that make the following function differentiable?

Answer 1

See explanation.

Explanation:

For this function to be differentiable at x=1, two things must be true:

1) `(1)^3 = m(1)+b`
2) The derivative of `x^3` and `mx+b` must be equal at `x=1`

If condition 1 is not met, the function is discontinuous . If condition 2 is not met, then our function experiences an entirely abrupt change in its rate of increase/decrease at that point, and thus the function is not differentiable there.

First, we look at condition 1:

`x^3 = mx+b, x=1 -> 1^3 = m+b -> m+b=1`

Thus, our first condition indicates that so long as the sum of m and b is equal to 1, the continuity condition is satisfied.

Condition 2: First, differentiate both parts of the piecewise function. The derivative of the first part will be the left hand derivative, and the derivative of the second will be the right hand derivative.

`(df)/dx = 3x^2` for `x<=1`, and `(df)/dx = m` for `x>1`

Thus, we must have `3x^2 = m`

And at x=1...

`3(1)^2 = m -> 3=m`

Now substituting our value for m into our analysis of condition 1:

`m+b=1 -> 3+b = 1 -> b = -2`

The function is differentiable at `x=1` if and only if `m=3, b=-2`