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

Let g(x) = x^2 and h(x) = mx+b
The piecewise function f(x) is basically a combination of g(x) and h(x) depending on what x is. 

If x is equal to 3 or smaller, then f(x) = g(x) = x^2
If x is larger than 3, then f(x) = h(x) = mx+b

Plug in x = 3 into g(x)
g(x) = x^2
g(3) = 3^2
g(3) = 9
And do the same for h(x)
h(x) = mx+b
h(3) = m*3+b
h(3) = 3m+b

In order for f(x) to be differentiable at x = 3, the two functions g(x) and h(x) must meet up at this x value. The function f(x) must be continuous at x = 3. In other words, g(x) = h(x) must be true when x = 3

So we can equate the two function outputs
g(3) = h(3)
9 = 3m+b
Solve for b to get
b = 9-3m


Now differentiate each function g(x) and h(x) with respect to x. Plug in x = 3 after you differentiated
g(x) = x^2
g ' (x) = 2x
g ' (3) = 2*3
g ' (3) = 6
h(x) = mx+b
h ' (x) = m
h ' (3) = m

If f(x) is differentiable at x = 3, then f ' (x) must be continuous at x = 3

This means,
g ' (x) = h ' (x)
g ' (3) = h ' (3)
6 = m
m = 6

Now use this value of m to find b
b = 9 - 3m
b = 9 - 3*6
b = 9 - 18
b = -9


In summary, we found that
m = 6
b = -9
which are the values needed to make f(x) differentiable

We need more info then that

6/9 i need more characters in my answer