Find the values of a and b such that f(x) is differentiable everywhere and compute f'(x)

Question

Let { f(x)={ e-1 / x^2 + x + 3, if x>0,

sin(ax) + b, if x ≤ 0.

(a) Find the values of a and b such that f(x) is differentiable everywhere and compute f'(x).

(b) Is the function f'(x) in part (a) differentiable everywhere? Compute f"(x) where it is defined.

I spent a few hours and I still can't figure out how to solve it..

Any help is appreciated.

Answer 1

I'm guessing the function looks like this:

If a piecewise function is differentiable everywhere, it must be differentiable at the boundary between the "pieces". That means it must be continuous there (the function value must be the same at x = 0 for the two pieces) and the function must be "smooth" there (no sharp points) which means the slope of the function (aka derivative) must be the same vale at x = 0 for the two pieces.

To make the function value the same, plug in x = 0 to the top "piece" to get 1/(3e) then set the bottom "piece" equal to 1/(3e). Since the since of zero is zero, then b = 1/(3e)

To make the derivative the same value, first take the derivative of both "pieces"

For the top "piece":

f(x) = e^-1/(x² + x + 3) you can re-write this as (1/e)(x² + x + 3)^-1 then use the power rule combined with the chain rule:

f ' (x) = (-1)(1/e)(x² + x + 3)^-2(2x + 1)

Then plug in x = 0:

f ' (0) = (-1)(1/e)(0² + 0 + 3)^-2(2(0) + 1) = -(1/e)(1/3²) = -1/(9e)

Now take the derivative of the lower "piece":

f(x) = sin(ax) + b

f '(x) = acos(ax)

Then plug in x = 0 to get:

f '(x) = acos(a(0)) = a•1 = a

Now set the two equal:

-1/(9e) = a

To determine if the function is differentiable everywhere, first consider that a sine function IS differentiable everywhere. The rational function is problematic only if the denominator is equal to zero, but solving the quadratic shows that it has no real solutions therefore is also differentiable everywhere.

To take the second derivative, the "bottom" piece is simple f '' (x) = -a²sin(ax). The "top" piece requires the use of the quotient rule. I'll leave that up to you.