If this function is differentials at x=1/2, what is a+b?

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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

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