help me find the absolute maximum!!!

Test Preparation
1

<p>f(x)= ((x^2) + 1)e^(-x) for -4<_ x <_ 4</p>

<p>a) for whay value of x does f reach its absolute maximum? justify your answer</p>

<p>b) find the x coordinates of all points of inflection of f. justify your answer</p>

2

<p>df(x)/dx = (2x)e^(-x)-((x^2)+1)e^(-x) = (-x^2+2x-1)e^(-x) = -((x-1)^2)/e^x</p>

<p>which is always nonpositive and is 0 only for x = 1</p>

<p>however f(1) is not any kind of min or max, because df(x)/dx does not change signs around 1. therefore over [-4,4] f(x) is decreasing, leaving the global max to be f(-4) = (16+1)e^4 = 17e^4</p>

<hr>

<p>f"(x) = -[2(x-1)e^x-((x-1)^2)e^x ] / e^(2x) =
= [-x^2+4x-3]/e^x
Critical points for inflection points when -x^2+4x-3 = 0, i.e. x1=1, x2=3
It's easy to chack that on different sides of these points f"(x) has different signs so 1 and 3 are inflection points.</p>

3

<p>i hope this isnt on the SAT...</p>

4

<p>No it's not.</p>

5

<p>how did you check to make sure that x=1,3 are inflection pts?</p>

6

<p>This is calculus level stuff, definitely not on the SAT's. How do you check? I don't know, graph it maybe.</p>

7

<p>inflection points are where the concavity changes...</p>

<p>so take the second derivative of f(x), and set it equal to zero. then take those numbers and do a sign chart to see where it changes signs. this is also how you tell if a graph is concave up or concave down.</p>

<p>good luck on curve sketching! if you have any other calc questions feel free to ask, i'll probably have an idea. I did well last year and am interning in two calc I classes and I am taking Calc II, so I see this stuff all the time...</p>

8

<p>haha sorry to post again, but if by checking you just mean verifying, all you have to say that the second derivative changes from positive to negative or vice versa, so therefore the concavity changes at that point.</p>