Finding inflection point

Question

I have a function given where I need to find the inflection point: $$y = (x^2 + 3\frac 34)\sqrt{x}$$ The first derivative is: $$y'= 2x\sqrt{x} + (x^2 + 3\frac 34)\frac{1}{2\sqrt{x}}$$ The second derivative is: \begin{align} y'' & = \frac{1}{\sqrt{x}} + \frac{2x}{2\sqrt{x}} + \left(x^2 + 3\frac 34\right)\left(-\frac{1}{4x}\right)\\ & = \frac{1}{\sqrt{x}} + \frac{x}{\sqrt{x}} - \frac{x^2 + 3\frac 34}{4x} \\ & = \frac{1}{\sqrt{x}} + \sqrt{x} - \frac{x^2 + 3\frac 34}{4x}\\ \end{align} To find the inflection point I need to equal the second derivative to $0$: $$0 = \frac{1}{\sqrt{x}} + \sqrt{x} - \frac{x^2 + 3\frac 34}{4x}$$ Now I don't know how to continue.

Answer 1

Your first derivative is correct but your second derivative is incorrect. I'll simplify your first derivative to make the second derivative less error-prone.

$$\begin{align} y' &= 2x\sqrt{x} + (x^2 + 3\frac 34)\frac{1}{2\sqrt{x}} \\[2 ex] &= 2x^{3/2}+\frac 12(x^2+\frac{15}4)x^{-1/2} \\[2 ex] &= 2x^{3/2}+\frac 12x^{3/2}+\frac{15}8x^{-1/2} \\[2 ex] &= \frac 52x^{3/2}+\frac{15}8x^{-1/2} \\[4 ex] y''&= \frac{15}4x^{1/2}-\frac{15}{16}x^{-3/2} \\[2 ex] &= \frac{15}4\sqrt x-\frac{15}{16x\sqrt x} \end{align}$$

(I checked the first and second derivatives on my graphing calculator and they check out.)

Set that expression for $y''$ equal to zero and multiply both sides by $x\sqrt x$ and you get a solvable equation.