The "concavity" of the function comes from the second derivative, where there is a critical point.
If the second derivative is positive at a point where the first derivative is zero, the function is "concave upward" there. If it is negative, it is concave downward.
FOr the function f(x)=x^2e^x
Find the critical points, increasing decreasing, first and second derivative and concavity
I am having trouble finding the concavity >_<
First derivative: xe^x(x+2)
Second : e^x(x^2+4x+2)
Critical points from first derivate:
x=0,-2
Increasing: (-infinity,0) (-2, infinity)
Decreasing: (0,-2)
I can't figure out how to do the concavity part of it.
2 answers
Increasing: (-infinity,0) (-2, infinity)
Decreasing: (0,-2)
This does not make sense. Coming from -oo you get to x = -2 before you get to x = 0
Now if I put in a large - number, say -100 for x
I get (-100)^2 e^-100= 10^4 *3.72*10^-44
which sure looks like 0
If I put in x = -2 I get
(-2)^2 e^-2 = .54
so it increases from x = -oo to x = 0
Now at x = 0, we know f(x) is zero, so it is decreasing from x = -2 to x = 0
We can be sure that it then heads up to the right so increasing from x = 0 to x --> +oo
Well, I suppose concave means curving upward. That would happen where the second derivative is +
Now where is the second derivative zero?
x^2+4x + 2 = 0
x = [ -4 +/- sqrt(16 -8) ] /2
= -2 +/- .5 sqrt 8
= -2 +/- sqrt 2
so the second derivative may change sign at -3.41 and at -.586
in other words to the left and right of where the function is level at (-2,.54)
To the left of -3.41 ,say x = -5, the second derivative is +, so curving up, concave from -oo to -3.41
at x = -2, the second derivative is negative, so curves down from -3.41 to - .586
Then at x = 0 the second derivative is + again so it starts curving up at x = -.586 and continues curving up from then on
Decreasing: (0,-2)
This does not make sense. Coming from -oo you get to x = -2 before you get to x = 0
Now if I put in a large - number, say -100 for x
I get (-100)^2 e^-100= 10^4 *3.72*10^-44
which sure looks like 0
If I put in x = -2 I get
(-2)^2 e^-2 = .54
so it increases from x = -oo to x = 0
Now at x = 0, we know f(x) is zero, so it is decreasing from x = -2 to x = 0
We can be sure that it then heads up to the right so increasing from x = 0 to x --> +oo
Well, I suppose concave means curving upward. That would happen where the second derivative is +
Now where is the second derivative zero?
x^2+4x + 2 = 0
x = [ -4 +/- sqrt(16 -8) ] /2
= -2 +/- .5 sqrt 8
= -2 +/- sqrt 2
so the second derivative may change sign at -3.41 and at -.586
in other words to the left and right of where the function is level at (-2,.54)
To the left of -3.41 ,say x = -5, the second derivative is +, so curving up, concave from -oo to -3.41
at x = -2, the second derivative is negative, so curves down from -3.41 to - .586
Then at x = 0 the second derivative is + again so it starts curving up at x = -.586 and continues curving up from then on
0 answers