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

Solution:

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

Thus, e^2b = 3 + b

The roots of the function,

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

First, prove it must have 2 roots.

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

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

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

By considering asymptotic behaviour, the g(b) goes to infinity as b goes to +infinity or -infinity.

Minimum function = g'(b) = 0

⇒ 2e^2b -1 = 0

⇒ 2e^2b = 1

⇒ b = ln(1/2)/2

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

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

We know that there exists b where g(b)<0 and g(b)>0 for b sufficiently large and b sufficiently small.

g(b) must have 2 roots.

Evaluating g(-2), we get

g(-2) = e^-4 - 3 + 2

g(-2) = e^-4 - 1

g(-2) < 0

Thus, the inequality follows e^-4<1.

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

Therefore, the values of b for which f is continuous at x = 2 is b = ln(1/2)/2.

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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

Summary:

Let f be the function defined.

f(x) = {e^bx for x ≤ 2
{1.5x + b for x ≥ 2

Function f is continuous at x = 2 for the value of b = ln(1/2)/2.