# curve sketching

• Jun 17th 2008, 08:26 AM
JMV
curve sketching
if someone could show me all the steps to curve sketch y=3xe^-x that would be greatly appreciated!(Doh)
• Jun 17th 2008, 09:01 AM
Reckoner
Quote:

Originally Posted by JMV
if someone could show me all the steps to curve sketch y=3xe^-x that would be greatly appreciated!(Doh)

My suggestions:

-Try to determine the domain and range of the function

-Find the $\displaystyle x$- and $\displaystyle y$- intercepts (if any)

-Find the derivative and determine where the function is increasing and decreasing

-Find the critical numbers of $\displaystyle f$ and use the first derivative test to find relative extrema

-Find second derivative and determine concavity and inflection points

-Sketch a few other points as needed

In this case, we have:

-Domain: $\displaystyle f(x)$ is defined for all real $\displaystyle x$, so our domain is the entire real line

-Intercepts:

$\displaystyle 3xe^{-x} = \frac{3x}{e^x} = 0$

$\displaystyle \Rightarrow 3x = 0\Rightarrow x = 0$

so our only intercept is at $\displaystyle (0, 0)$.

-Increasing/decreasing and extrema:

$\displaystyle f'(x) = (3x)\left(-e^{-x}\right) + e^{-x}(3) = 3e^{-x}(1 - x)$

Note that $\displaystyle f'$ is defined everywhere. Setting $\displaystyle f'(x) = 0$, we have

$\displaystyle 3e^{-x}(1 - x) = 0\Rightarrow3(1 - x) = 0\Rightarrow x = 1$

and $\displaystyle f$ has one critical point, at $\displaystyle \left(1,\;\frac3e\right)$. By testing $\displaystyle f'$ on the intervals $\displaystyle (-\infty,\;1)$ and $\displaystyle (1,\;\infty)$, you should be able to determine that $\displaystyle f$ is increasing over the first interval and decreasing over the second, so this point is a relative maximum (and in fact it is an absolute maximum).

-Concavity and inflection points:

Differentiate: $\displaystyle f''(x) = 3e^{-x}(-1) + (1 - x)\left(-3e^{-x}\right) = 3e^{-x}(x - 2)$

So we have a possible inflection point at $\displaystyle x = 2$. Testing the intervals, you should find that $\displaystyle f''(x) < 0\;\forall x\in(-\infty,\;2)$ (concave down) and $\displaystyle f''(x) > 0\;\forall x\in(2,\;\infty)$ (concave up). Thus $\displaystyle \left(2,\;\frac6{e^2}\right)$ is an inflection point.

Plot the appropriate points, and use the concavity and intervals of increasing/decreasing values to sketch the graph.
• Jun 17th 2008, 10:03 AM
Soroban
Hello, JMV!

Quote:

Graph: .$\displaystyle y \:=\:3xe^-x$
The only intercept is the origin: (0, 0).

The function is: .$\displaystyle y \:=\:\frac{3x}{e^x}$

Since $\displaystyle e^x \neq 0$, there are no vertical asymptotes.

Since $\displaystyle \lim_{x\to\infty}\frac{3x}{e^x} \:=\:0$, the horizontal asymptote is: .$\displaystyle y \:=\:0$ (x-axis)

First derivative: .$\displaystyle y' \:=\:-3xe^{-x} + 3x^{-x} \:=\:3e^{-x}(1-x) \:=\:\frac{3(1-x)}{e^x}$

We see that: for $\displaystyle x = 1$, horizontal tangent (critical point).
. . For $\displaystyle x < 1,\:y' > 0$ ... graph is rising (up to the critical point)
. . For $\displaystyle x > 1,\:y' < 0$ ... graph is falling (down to the x-axis)

When $\displaystyle x = 1,\:y \:=\:\frac{3}{2} \:\approx\:1.1$ . . . We have the point (1, 1.1)
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We can examine the second derivative.

It simplifies to: .$\displaystyle y'' \:=\:\frac{3(x-2)}{e^{2x}}$

. . For $\displaystyle x < 2,\:y'' < 0$, concave down

. . At $\displaystyle x = 2,\:y'' = 0$, inflection point at about (2, 0.8)

. . For $\displaystyle x > 2,\:y'' > 0$, concave up

All of which agrees with my sketch . . . whew!

• Jun 17th 2008, 10:09 AM
Mathstud28
Quote:

Originally Posted by Soroban

When $\displaystyle x = 1,\:y \:=\:\frac{3}{2} \:\approx\:1.1$ . . . We have the point (1, 1.1)

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