On what interval is this function concave downward?

Question

f(x) = (x^2)(e^x)
I was able to get it down to (x^2)(e^x)+(e^x)(2x) but then I'm not sure what to do. Do I just plug in numbers until I can get it to negative? In which case it would be x < -1, without going into crazy decimals?

Thanks for any help.

Answer 1

Ok buddy im sure you have a graphing calculator...Simply graph the original function f(x)=(x^2)(e^x) on the calculator and see where the function hits the x-axis. I am sure you know what concave up and concave down look like so you can approximate the answer from there...HOWEVER if you really wish to prove it is concave down take the second derivative of f(x)=(x^2)(e^x) which by the way is e^x(x^2+4x+2). The zeros of this function are -3.41421 and 0. If this equation at any time is negative then the function is concave down...Hope this helps!

Answer 2

You need to find where the second derivative in negative.

First find the second derivative.

Use this site and put in x^2*e^x as the function.

http://www.calc101.com/webMathematica/derivatives.jsp

This will show you much better than I can both what the derivative is and how to find it.

You get the following:
e^x*(x^2+4x+2)

Now the e^x is always greater than zero, so you only need to worry about the polynomial. Solve for the roots of x^2+4x+2 using the quadratic formula, and you get:

-2-sqrt(2) and -2+sqrt(2)

Which are -3.414121 and -0.585786

Pick an easy point between the two, say -1 and plug it into the polynomial and you will see that it is negative at that point, so the original equation is concave down between those two roots.

Answer 3

You need the second derivative. Set the second derivative equal to zero and solve for x (you will need to factor out e^x). Solving for x gives you the critical points. Test numbers on the intervals between the points. When you get negative numbers for the interval you will have the intervals where the function in concave down.

Answer 4

To find when the function is concave down, we must find when it's second derivative is negative.

f(x)=(x^2)(e^x)
f'(x)=2x(e^x)+(x^2)(e^x)= (x^2+2x)(e^x)
f''(x)= (x^2+2x)(e^x)+(2x+2)(e^x) = (e^x)(x^2+4x+2)


e^x is always positive, so we only need to consider when x^2+4x+2.

descriminant= 4^2-4(2)=8, thus f''(x) has zeros when x=-2±√2

Think about the graph of f''(x), it is clear that it is negative only when the value of x is between its two roots (-2±√2), thus it is concave down on the interval (-2-√2,-2+√2).

Answer 5

examine to work out the position the second one by-product is decrease than 0. f(x) = x^2 e^x f'(x) = 2x e^x + x^2 e^x f''(x) = 2e^x + 2xe^x + 2xe^x + x^2 e^x f''(x)=2e^x +4xe^x + x^2 e^x to work out on what period this function is detrimental, it really is about (minus infinity, -0.00007) or more suitable about (minus infinity, 0) you need to graph the function.

Answer 6

f(x)=x^2e^x
f'(x)=x^2e^x+2xe^x
f''(x)=x^2e^x+4xe^x+2e^x
f''(x)=0 @ x= -2+(2)^1/2, -2-(2)^1/2
check values of x greater than -2+(2)^1/2

your f'(x) is not right.