Let f be the function defined. For what values of b is f continuous at x=2

Question

Let f be the function defined. For what values of b is f continuous at x=2

f(x)= {e^bx for x≤2

{1.5x + b for x≥2

1) 0.508 only

2) 0.647 only

3) -1.282 and 0.508

4) -2.998 and 0.647

I'd also like an explanation, if you can. Thanks!

Answer 1

The correct answer is 4), but I do not have an analytical solution.

This is my reasoning.

For f to be continuous at 2, the two definitions must agree for x = 2.

That is,

e^2b = 3 + b

So we are looking for roots of the function

g(b) = e^2b - 3 - b

(I) First we prove that it must have 2 roots.

Compute

g'(b) = 2e^2b - 1

g''(b) = 4e^2b

Since the second derivative g"(b) is always positive, the function g(b) is convex.

And it goes to infinity as b goes to +infinity or -infinity.

(That is seen easily by considering its asymptotic behavior.)

We compute the minimum of the function by setting its derivative to 0.

2e^2b - 1 = 0

b = ln(1/2)/2

Now we check that g(b) is negative at this minimum.

g(ln(1/2)/2) = 1/2 - 3 - ln(1/2)/2 < 0

Now that we know that there exists b where g(b) < 0 and that g(b) > 0 for b sufficiently

large and b sufficiently small, we know that g(b) must have 2 roots.

(II) This leaves us with option 3) or 4).

Now we prove that it must be 4) by evaluating

g(-2) = e^-4 - 3 + 2 = e^-4 - 1 < 0

The last inequality follows from the fact that e^-4 < 1.

Since g(-2) < 0, it must have a root < -2.

That leaves only option 4)