Finding inflection points

[Math_Junkie]

Find the inflection point of the following equation:

I know you find the second derivative and equate it to 0.. but I can't seem to solve the second derivative for this. Showing my work for this would turn out into a mess. Is anyone else able to find an inflection point for this?

 

[Sohcahtoa]

This function looks nasty, but after you compute f '(x) using the product rule, factor out 7e[sup:17q9acf5]-x[/sup:17q9acf5]. You will have a f '(x)=7e[sup:17q9acf5]-x[/sup:17q9acf5](g(x)), where g(x) involves x[sup:17q9acf5]1/2[/sup:17q9acf5] and x[sup:17q9acf5](-1/2)[/sup:17q9acf5]. Now when you take the second derivative of f(x), you use the product rule again, factor the 7e[sup:17q9acf5]-x[/sup:17q9acf5] out again, and you have f"(x)=7e[sup:17q9acf5]-x[/sup:17q9acf5]h(x), where h(x) has x[sup:17q9acf5](-3/2)[/sup:17q9acf5], x[sup:17q9acf5](-1/2)[/sup:17q9acf5], and x[sup:17q9acf5](1/2)[/sup:17q9acf5] terms.
Then recall that the domain of f(x) is all non-negative numbers and that e[sup:17q9acf5]-x[/sup:17q9acf5]>0 always, so you just have to set h(x) = 0 for all possible non-negative values of x to find the inflection point(s). Use a common denominator to get h(x)=r(x)/d(x) and set r(x) = 0 to find your inflection point. r(x) will be a quadratic function in x with one root in the domain of f(x).

 

[soroban]

Hello, Math_Junkie!

Thought I'd give it a try . . .

$\displaystyle \text{Find the inflection points of: }\;f(x) \:=\:7\sqrt{x}\,e^{-x}$

$\displaystyle \text{Note that: }\:x \geq 0$


$\displaystyle \text{We have: }\;f(x) \;=\;7\cdot\frac{x^{\frac{1}{2}}}{e^x}$

$\displaystyle f'(x) \;=\;7\cdot\frac{e^x\cdot\frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}\cdot e^x}{e^{2x}} \;=\;7\cdot \frac{e^x\left(\frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}\right)} {e^{2x}} \;=\;\frac{7}{2}\cdot\frac{\frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}}{e^x}$

$\displaystyle \text{Multiply by }\frac{2x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}\!:\quad f'(x) \;=\;7\cdot\frac{1-2x}{2x^{\frac{1}{2}}e^x} \;=\;\frac{7}{2}\cdot\frac{1-2x}{x^{\frac{1}{2}}e^x}$


$\displaystyle f''(x) \;=\;\frac{7}{2}\cdot\frac{x^{\frac{1}{2}} e^x(-2) - (1-2x)\left(x^{\frac{1}{2}}e^x + \frac{1}{2}x^{-\frac{1}{2}}e^x\right)} {xe^{2x}}$

. . . . .$\displaystyle =\; \frac{7}{2}\cdot \frac{-2x^{\frac{1}{2}}e^x - x^{\frac{1}{2}}x^x - \frac{1}{2}x^{-\frac{1}{2}}e^x + 2x^{\frac{3}{2}}e^x + x^{\frac{1}{2}}e^x} {xe^{2x}}$

. . . . .$\displaystyle =\; \frac{7}{2}\cdot \frac{2x^{\frac{3}{2}}e^x - 2x^{\frac{1}{2}}e^x - \frac{1}{2}x^{-\frac{1}{2}}e^x}{xe^{2x}}$

. . . . .$\displaystyle =\; \frac{7}{2}\cdot \frac{e^x\left(2x^{\frac{3}{2}} - 2x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}\right)}{xe^{2x}} \;=\; \frac{7}{2}\cdot \frac{2x^{\frac{3}{2}} - 2x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}} {xe^x}$

$\displaystyle \text{Multiply by }\frac{2x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}\!:\quad f''(x) \;=\;\frac{7}{2}\cdot\frac{4x^2 - 4x - 1}{2x^{\frac{3}{2}}e^x}$

$\displaystyle \text{Hence: }\;f''(x) \;=\;\frac{7}{4}\cdot\frac{4x^2 -4x-1}{x^{\frac{3}{2}}e^x}$


$\displaystyle \text{Inflection points occur where }f''(x) \,=\,0 \quad\Rightarrow\quad 4x^2 - 4x - 1 \:=\:0$

$\displaystyle \text{Quadratic Formula: }\:x \;=\;\frac{4 \pm\sqrt{16+16}}{8} \;=\;\frac{4\pm\sqrt{32}}{8} \;=\;\frac{4\pm4\sqrt{2}}{8} \;=\;\frac{1 \pm\sqrt{2}}{2}$

$\displaystyle \text{But }x \,=\,\frac{1 - \sqrt{2}}{2}\,\text{ is not in the domain.}$

$\displaystyle \text{Therefore, the only inflection is at: }\:x \,=\,\frac{1+\sqrt{2}}{2}$

 

[Math_Junkie]

Thanks so much to the both of you. I made some small errors when finding the second derivative.

Thanks again!