Finding a Point of Inflexion for f(x)

[cmkluza]

Homework Statement

The function $f$ is defined on the domain $x≥0$ by $f(x) = \frac{x^2}{e^x}$.
(a) Find the maximum value of $f(x)$, and justify that it is a maximum.
(b) Find the $x$ coordinates of the points of inflexion on the graph of $f$.

I believe that I did (a) correctly, but (b) is where I got stuck.

Homework Equations

Point of Inflexion Criteria:

• Is 0 at $f''(x)$
• Changes Sign at $f''(x) = 0$

I believe that's all that's really relevant, and of course differentiation, power/quotient rules.

The Attempt at a Solution

(a)
$f(x) = \frac{x^2}{e^x}$
$f'(x) = \frac{2xe^x - x^2e^x}{e^{2x}}$
∴ $2xe^x - x^2e^x = 0$
$2xe^x = x^2e^x$
$2x = x^2$
$x=2$
Since the exponentially growing $e^x$ will increase at a much greater pace than the $x^2$ term, in $f(x) = \frac{x^2}{e^x}$, there can be no other maxima.

(b)
This is the part I had trouble on. I figured the second derivative to be:
$f''(x) = \frac{3x^2-4x+2}{e^x}$
∴ $3x^2-4x+2 = 0$
$x = \frac{4±\sqrt{40}}{6} \rightarrow x ≥ 0$
∴ $x = \frac{4+\sqrt{40}}{6}$

I've probably made some mistake somewhere in my algebra, or more likely that annoying differentiation, but when I plug values on either side of that resulting x back into $f''(x)$, there is no change in sign. Am I missing something? Have I made a mistake somewhere in my algebra or calculus? I don't believe there could be any other points that equal zero in the equation I've generated, so where'd I go wrong?

Thanks to anyone who can help me with this problem!

 

[SteamKing]

Homework Statement

The function $f$ is defined on the domain $x≥0$ by $f(x) = \frac{x^2}{e^x}$.
(a) Find the maximum value of $f(x)$, and justify that it is a maximum.
(b) Find the $x$ coordinates of the points of inflexion on the graph of $f$.

I believe that I did (a) correctly, but (b) is where I got stuck.

Homework Equations

Point of Inflexion Criteria:

• Is 0 at $f''(x)$
• Changes Sign at $f''(x) = 0$

I believe that's all that's really relevant, and of course differentiation, power/quotient rules.

The Attempt at a Solution

(a)
$f(x) = \frac{x^2}{e^x}$
$f'(x) = \frac{2xe^x - x^2e^x}{e^{2x}}$
∴ $2xe^x - x^2e^x = 0$
$2xe^x = x^2e^x$
$2x = x^2$
$x=2$
Since the exponentially growing $e^x$ will increase at a much greater pace than the $x^2$ term, in $f(x) = \frac{x^2}{e^x}$, there can be no other maxima.

(b)
This is the part I had trouble on. I figured the second derivative to be:
$f''(x) = \frac{3x^2-4x+2}{e^x}$
∴ $3x^2-4x+2 = 0$
$x = \frac{4±\sqrt{40}}{6} \rightarrow x ≥ 0$
∴ $x = \frac{4+\sqrt{40}}{6}$

I've probably made some mistake somewhere in my algebra, or more likely that annoying differentiation, but when I plug values on either side of that resulting x back into $f''(x)$, there is no change in sign. Am I missing something? Have I made a mistake somewhere in my algebra or calculus? I don't believe there could be any other points that equal zero in the equation I've generated, so where'd I go wrong?

Thanks to anyone who can help me with this problem!

Using the quotient rule to take the first and second derivatives is the hard way to go.

Re-write f(x) = x² * e^-x instead and take both derivatives again. This can serve as a check on your original work, which I don't think is entirely correct, especially f"(x). Note: f'(x) can be simplified from your result.

 

[cmkluza]

Using the quotient rule to take the first and second derivatives is the hard way to go.

Re-write f(x) = x² * e^-x instead and take both derivatives again. This can serve as a check on your original work, which I don't think is entirely correct, especially f"(x). Note: f'(x) can be simplified from your result.

I've just double checked. First derivative is (presumably) correct, and simplified down to:
$f'(x)=\frac{2x-x^2}{e^x}$
Second derivative I did indeed get a different result for:
$f''(x)=\frac{x^2-3x}{e^x}$

After that I figured $x=3$ and it all worked out. Thanks for your suggestion! I don't know why I didn't go for the power rule in the first place, but anyways, thanks for your assistance!

 

[Ray Vickson]

Homework Statement

The function $f$ is defined on the domain $x≥0$ by $f(x) = \frac{x^2}{e^x}$.
(a) Find the maximum value of $f(x)$, and justify that it is a maximum.
(b) Find the $x$ coordinates of the points of inflexion on the graph of $f$.

I believe that I did (a) correctly, but (b) is where I got stuck.

Homework Equations

Point of Inflexion Criteria:

• Is 0 at $f''(x)$
• Changes Sign at $f''(x) = 0$

I believe that's all that's really relevant, and of course differentiation, power/quotient rules.

The Attempt at a Solution

(a)
$f(x) = \frac{x^2}{e^x}$
$f'(x) = \frac{2xe^x - x^2e^x}{e^{2x}}$
∴ $2xe^x - x^2e^x = 0$
$2xe^x = x^2e^x$
$2x = x^2$
$x=2$
Since the exponentially growing $e^x$ will increase at a much greater pace than the $x^2$ term, in $f(x) = \frac{x^2}{e^x}$, there can be no other maxima.

(b)
This is the part I had trouble on. I figured the second derivative to be:
$f''(x) = \frac{3x^2-4x+2}{e^x}$
∴ $3x^2-4x+2 = 0$
$x = \frac{4±\sqrt{40}}{6} \rightarrow x ≥ 0$
∴ $x = \frac{4+\sqrt{40}}{6}$

I've probably made some mistake somewhere in my algebra, or more likely that annoying differentiation, but when I plug values on either side of that resulting x back into $f''(x)$, there is no change in sign. Am I missing something? Have I made a mistake somewhere in my algebra or calculus? I don't believe there could be any other points that equal zero in the equation I've generated, so where'd I go wrong?

Thanks to anyone who can help me with this problem!

Please: never again write $1/e^x$; always convert it to $e^{-x}$. Believe it or not, that will simplify your life a lot!