This is a curve sketching/analysis question I have with a graph involving e^x because I'm unsure about how the graph of this changes, etc.
Link to the image in case you can't see it:
http://img243.imageshack.us/img243/3343/1812ge5.png
This is a curve sketching/analysis question I have with a graph involving e^x because I'm unsure about how the graph of this changes, etc.
Link to the image in case you can't see it:
http://img243.imageshack.us/img243/3343/1812ge5.png
What do you mean when you say "Fully analyze the graph"? Are you looking for the critical points, such as maximums/minimums, points of inflection, increasing/decreasing slopes, intervals of upward/downward concavity?
Question 12 is clear, so I'll do that for now until you specify what you need for 10.
h(x) = sqrt(lnx)
h'(x) = 1/2*1/sqrt(lnx)*1/x = 1/(2x*lnx)
h'(e) = 1/(2*e*lne) = 1/(2e)
I leave "fully analyzing" the function to you, I'll go to the meat of the matter.
We have the largest rectangle when the area of the rectangle is largest, so let's maximize the area.
What is our formula for the area? length times width of course. so we have,
A = 2x*e^(-2x^2)
=> A' = 2e^(-2x^2) + -8x^2*e^(-2x^2)
For max A, set A' = 0
=> 2e^(-2x^2) + -8x^2*e^(-2x^2) = 0
=> 2e^(-2x^2)(1 - 4x^2) = 0
=> 2e^(-2x^2)(1 + 2x)(1 - 2x) = 0
=> 2e^(-2x^2) = 0 ......impossible
or (1 + 2x) = 0 => x = -1/2
or (1 - 2x) = 0 => x = 1/2
So we have the largest area for the points (-1/2,0) and (1/2 , 0)
(Do you need help with the "fully analyzing" part?)
Ah, i'm kinda bored. let's see how far we get with these guys.
let y = f(x) = e^(-2x^2)
now f(-x) = e^(-2(-x)^2) = e^(-2x^2) = f(x)
since f(x) = f(-x), y is symmetric about the y-axis
For x-intercept, set y = 0b. find intercepts
=> e^(-2x^2) = 0
no solution, thus there is no x-intercept
For y-intercept, set x = 0
=> y = e^(-2(0)^2) = e^0 = 1
so we have one y-intercept, at y = 1
f(x) is increasing in an interval if f'(x)>= 0 for all x in that intervalc. find intervals of increase or decrease
f(x) is decreasing in an interval if f'(x)<= 0 for all x in that interval
now f'(x) = -4x*e^(-2x^2)
e^(-2x^2) is always positive, so the sign of f'(x) depends on the -4x
-4x > 0 if x < 0 .......so the function is increasing when this happens
-4x < 0 if x > 0 .......so the function is decreasing when this happens
note that where -4x = 0, f'(x) = 0 and so we have a critical point. further analysis will show us that this is a local maximum.
so f(x) is increasing on (-infinity, 0) and decreasing on (0,infinity) it is at rest at x = 0
we have the possibility of inflection points where the second derivative = 0d. find inflection points
now f''(x) = 16x^2 *e^(-2x^2)
f''(x) = 0 when x = 0 .......so this is a possible inflection point. let's make sure. if the sign of the derivative on both sides of the point is the same, it is an inflection point, otherwise, it is a max or min. take two numbers, one to the right of x = 0 and one to the left. say,
x1 = -1
x2 = 1
f ' (-1) > 0
f ' (1) < 0
since we are increasing on the left and decreasing on the right, it means we have a maximum, not an inflection point. there are no inflection points, since the only possibility turned out to be a maximum.
as you saw, there is one critical point, at x = 0, it is a maximum and hence we have concave down. so the function is concave down everywhere, that is on the interval (-infinity, infinity)e. find concavity
i did that when i was solving the largest rectangle problem, see the post above (would you know how to draw a rough sketch if you had to do it by hand?)f. sketch the curve