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.

  • 2
    $\begingroup$ 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. $\endgroup$ – randomgirl Jan 20 '16 at 4:41
  • $\begingroup$ 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. $\endgroup$ – T. Eskin Jan 20 '16 at 4:46
  • $\begingroup$ The graph intersects the y-axis at the point (0,2), therefore, maxima is 2 at x=0. I graphed the equation. $\endgroup$ – Calc Jan 20 '16 at 4:49
  • $\begingroup$ What graph are you looking at? Just insert $x=0$ into $y=e^{-5x^2}$ and you will see y is not $2$ when $x=0$. $\endgroup$ – randomgirl Jan 20 '16 at 4:50
  • $\begingroup$ wolframalpha.com/input/?i=y%3De%5E%28-5x%5E2%29 graph should look like this $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.