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Work Shown:
Let
g(x) = ax^3 + 2
h(x) = bx^2 + 1
Plug x = 1/2 into each function
g(x) = ax^3 + 2
g(1/2) = a(1/2)^3 + 2
g(1/2) = (1/8)a + 2
and
h(x) = bx^2 + 1
h(1/2) = b(1/2)^2 + 1
h(1/2) = (1/4)b + 1
In order for f(x) to be differentiable at x = 1/2, the function f(x) must be continuous at x = 1/2. If we had a jump discontinuity here, then there is no way the function is differentiable here.
f(x) is continuous at x = 1/2 when g(1/2) = h(1/2).
This means,
g(1/2) = h(1/2)
(1/8)a + 2 = (1/4)b + 1
8*[(1/8)a + 2] = 8*[(1/4)b + 1]
a+16 = 2b+8
a = 2b+8-16
a = 2b-8
We'll keep this in mind for later.
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Now let's compute the derivatives
g(x) = ax^3 + 2
g ' (x) = 3ax^2
h(x) = bx^2 + 1
h ' (x) = 2bx
Now plug in x = 1/2
g ' (x) = 3ax^2
g ' (1/2) = 3a(1/2)^2
g ' (1/2) = (3/4)a
and
h ' (x) = 2bx
h ' (1/2) = 2b(1/2)
h ' (1/2) = b
Equate the two items. We do so because we want f(x) to be differentiable at x = 1/2, so that means g ' (1/2) = h ' (1/2)
So,
g ' (1/2) = h ' (1/2)
(3/4)a = b
3a = 4b
a = (4b)/3
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Earlier we found a = 2b-8.
Since both a = 2b-8 and a = (4b)/3, this means we can equate the right hand sides to solve for b
2b-8 = (4b)/3
3*(2b-8) = 3*(4b)/3
6b-24 = 4b
6b-4b = 24
2b = 24
b = 24/2
b = 12
Which leads to
a = 4b/3
a = 4*12/3
a = 48/3
a = 16
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The piecewise function
[tex]f(x) = \begin{cases}ax^3+2, \ \ x < \frac{1}{2}\\bx^2+1, \ \ x \ge \frac{1}{2}\end{cases}[/tex]
will update to
[tex]f(x) = \begin{cases}16x^3+2, \ \ x < \frac{1}{2}\\12x^2+1, \ \ x \ge \frac{1}{2}\end{cases}[/tex]
To verify f(x) is differentiable at x = 1/2, you should check to see if both pieces evaluate to the same output when you plug in x = 1/2. In other words, you should check to see if g(1/2) = h(1/2).
Also, you should check to see if g ' (1/2) = h ' (1/2) so that f(x) is fully differentiable here. I'll let you do the check portions.
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After getting our a and b values, we can finally say that
a+b = 16+12 = 28
