Thread: Finding equation given inflection point

1. Finding equation given inflection point

$\displaystyle f(x)=\sqrt{x+1}+\frac{b}{x}$

Find $\displaystyle b$ such that $\displaystyle f(x)$ has a point of inflection at $\displaystyle x=3$

Answer in the back has it at $\displaystyle \frac{27}{64}$, but I'm not sure how to do it.

2. For a point of inflection, $\displaystyle\frac{d^2y}{dx^2}=0$

3. Originally Posted by Quacky
For a point of inflection, $\displaystyle\frac{d^2y}{dx^2}=0$
Meaning I'd have to solve $\displaystyle f''(3)=0$ right? That's what I tried, but I got a completely different answer.

4. Let me have a go.

$f(x)=\displaystyle (x+1)^{\frac{1}{2}}+\frac{b}{x}$
$
f'(x)=\displaystyle\frac{1}{2}(x+1)^{-\frac{1}{2}}-\frac{b}{x^2}$

$f''(x)=\displaystyle -\frac{1}{4}(x+1)^{\frac{-3}{2}}+\frac{2b}{x^3}$

When $x=3, f''(x)=0$

I carried on from here, and was able to get the correct answer of $\frac{27}{64}$, but I admit that I initially made a differentiation error myself.

5. Originally Posted by Quacky
Let me have a go.

$f(x)=\displaystyle (x+1)^{\frac{1}{2}}+\frac{b}{x}$
$
f'(x)=\displaystyle\frac{1}{2}(x+1)^{-\frac{1}{2}}-\frac{b}{x^2}$

$f''(x)=\displaystyle -\frac{1}{4}(x+1)^{\frac{-3}{2}}+\frac{2b}{x^3}$

When $x=3, f''(x)=0$

I carried on from here, and was able to get the correct answer of $\frac{27}{64}$, but I admit that I initially made a differentiation error myself.
Thanks a lot! These kind of questions really annoy me>.< Thanks again