# Exponential and Logarithmic Functions: $y=2e^{-5x^2}$

Use the function $$y=2e^{-5x^2}$$

I have already answered most of the questions, but I would really appreciate it if you would look over my answers and tell me if I am wrong and help me correct my mistakes. I mostly need help with e) and g)

a) State the domain:

I wrote domain as $x\in\mathbb R$.

b) Determine the intercepts, if any:

No x-intercept, and y-intercept at $(0,2)$.

c) Discuss the symmetry of the graph:

I said the graph is symmetric with respect to the y-axis because the function is even.

d) Find any asymptotes:

No vertical asymptote, horizontal asymptote at $y=0$.

e) Determine the intervals of increase and decrease:

I know I'm supposed to set the first derivative to zero then solve, but I get lost and I need help. How do I find the intervals of increase and decrease?

f) What is the maxima and/or minima:

Maxima is $2$ at $x=0$, and minima value does not exist.

g) Where is the curve concave upward or downward:

I know to set the second derivative to zero, but again, I get lost, I need help on this one.

h) Locate the points of inflection: Not sure if it's correct, but I got

$$\left(-\frac{1}{\sqrt{10}},\frac{2}{\sqrt e}\right) \text{ and } \left(\frac{1}{\sqrt {10}}, \frac{2}{\sqrt e}\right)$$

Thank-you in advance, your help is always appreciated.

• How did you get your "y-intercept"? You will also need to re-look at your max. I haven't check those other values towards the end there. – randomgirl Jan 20 '16 at 4:41
• Draw the graph to get an idea. It answers majority of the questions by itself. And see then how the first and second derivative tests give the same results. – T. Eskin Jan 20 '16 at 4:46
• The graph intersects the y-axis at the point (0,2), therefore, maxima is 2 at x=0. I graphed the equation. – Calc Jan 20 '16 at 4:49
• What graph are you looking at? Just insert $x=0$ into $y=e^{-5x^2}$ and you will see y is not $2$ when $x=0$. – randomgirl Jan 20 '16 at 4:50
• wolframalpha.com/input/?i=y%3De%5E%28-5x%5E2%29 graph should look like this – randomgirl Jan 20 '16 at 4:52

As for the increase/decrease thing, $f'$ will tell you this information. You already found when $f'=0$. Use test numbers on both sides of $x=0$ to see if $f'>0$ (this means $f$ is increasing) or if $f'<0$ (this means $f$ is decreasing). You can do something similar with the concavity thing. You found when $f''=0$ Just choose test numbers for each of the three intervals to see if $f''>0$ (this means $f$ is concave up) or if $f''<0$ (this means $f$ is concave down). If concavity switches, at the points where $f''=0$ then at these points you have an inflection point.