Cont. and Diff.

[johnjones]

I'm kind of stuck on this question.

Let f(x) = { x^2 x <= 1
ax^(1/2)+ b x > 1

(a) Find the condition on a and b so f will be continuous.
(b) find a and b so f will be differentiable.

My approach:
somehow, I think I have to make them both equal... but then, I have three unknown variables, a, x and b . I don't know how to solve for them :?: . Thx.

 

[pka]

Given that \(\displaystyle f(x) = \left\{ {x^2 } & {x \le 1} \\
{a\sqrt x + b} & {1 < x} \\
\ \right.\),

for continuity we must have \(\displaystyle \ {{\rm{lim}}}\limits_{x \to 1^ + } f(x) = \{{\rm{lim}}}\limits_{x \to 1^ - } f(x)\): \(\displaystyle a + b = 1\).

For it to have a derivative at 1, we must have: \(\displaystyle \frac{a}{{2\sqrt x }} = 2x\) or \(\displaystyle \frac{a}{2} = 2\).

 

[johnjones]

Given that \(\displaystyle f(x) = \left\{ {x^2 } & {x \le 1} \\
{a\sqrt x + b} & {1 < x} \\
\ \right.\),

for continuity we must have \(\displaystyle \ {{\rm{lim}}}\limits_{x \to 1^ + } f(x) = \{{\rm{lim}}}\limits_{x \to 1^ - } f(x)\): \(\displaystyle a + b = 1\).

For it to have a derivative at 1, we must have: \(\displaystyle \frac{a}{{2\sqrt x }} = 2x\) or \(\displaystyle \frac{a}{2} = 2\).

Where did 2 = a/2 come from?
I understand 2x = a/[2(x^0.5)].
When I solved for x, I got:

a = (2x)(2x^1/2) = 4x^(3/2).

 

[Gene]

You want
2x = a/[2(x^0.5)]
to be true at x=1 so just substitute 1 for x to get
2 = a/2