How do you determine whether the function #f(x) = x^2e^x# is concave up or concave down and its intervals?

Answer 1

You have to calculate inflection point(s), which mean find zeros of the second derivative.

#f''(x)=0#
#f(x)=x^2*e^(x)# #f'(x)=2*x*e^x+x^2*e^x# #f''(x)=2*e^x+2*x*e^x+2*x*e^x+x^2*e^x=# #=e^x(2+4*x+x^2)#
#f''(x)=0# #<=># #2*e^x(1+x+x^2)=0# Remember that #e^x# is always* positive #=> (1+x+x^2)=0# #x_(1,2) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}# #x_(1,2) = \frac{-4 \pm \sqrt{4^2 - 4*2*1}}{2*1}=# #=\frac{-4 \pm \sqrt{16 - 8}}{2}=# #=\frac{-4 \pm \sqrt{8}}{2}=# #=\frac{-4 \pm 2\sqrt{2}}{2}=# #=-2 \pm \sqrt{2}#
#x_1=-2 +\sqrt(2)=-0.59# #x_2=-2 -\sqrt(2)=-3.41# now you have 3 intervals: 1. #(-infty,x_2)# 2. #(x_2,x_1)# 3. #(x_1,infty)#
ask yourself if #f''(x)# is positive or negative on each interval: (choose a number of the interval and calculate the value, don't choose #x_1# or #x_2# because there is 0 as we calculated before). 1. #(-infty,-3.41)# #x=4, f''(-4)>0# 2. #(-3.41,-0.59)# #x=-1, f''(-1)<0# 3. #(-0.59,infty)# #x=0, f''(0)>0#
if #f''(x)>0# #=> f(x)# is convex if #f''(x)<0# #=> f(x)# is concave
convex interval: #(-infty,-3.41)uu(-0.59,infty)# concave interval: #(-3.41,-0.59)#
*#lim_(x->-infty) f''(x)=0# (we are looking for #abs(x) < infty#)
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To determine the concavity of the function ( f(x) = x^2e^x ), we need to find its second derivative and then analyze its sign.

First, find the first derivative of ( f(x) ) using the product rule:

[ f'(x) = (2x \cdot e^x) + (x^2 \cdot e^x) ]

Now, differentiate ( f'(x) ) with respect to ( x ) again to find the second derivative:

[ f''(x) = (2 \cdot e^x + 2x \cdot e^x) + (2x \cdot e^x + x^2 \cdot e^x) ]

[ f''(x) = (2 + 2x) \cdot e^x + (2x + x^2) \cdot e^x ]

[ f''(x) = (2 + 2x + 2x + x^2) \cdot e^x ]

[ f''(x) = (x^2 + 4x + 2) \cdot e^x ]

To determine the concavity of the function, analyze the sign of the second derivative. If ( f''(x) > 0 ), the function is concave up. If ( f''(x) < 0 ), the function is concave down.

The second derivative ( f''(x) = (x^2 + 4x + 2) \cdot e^x ) is always positive for all real values of ( x ). Therefore, the function ( f(x) = x^2e^x ) is concave up for all real values of ( x ). There are no intervals where it is concave down.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To determine the concavity of the function (f(x) = x^2e^x) and its intervals, you need to find the second derivative of the function and then analyze its sign.

  1. Find the first derivative of (f(x)) using the product rule: (f'(x) = 2xe^x + x^2e^x).
  2. Find the second derivative of (f(x)) using the product rule again: (f''(x) = (2e^x + 2xe^x) + (2xe^x + x^2e^x) = (2 + 2x)e^x + 2xe^x = (2 + 4x)e^x).
  3. To determine concavity, set (f''(x)) equal to zero and solve for (x): ((2 + 4x)e^x = 0). Since (e^x) is always positive, the only way for the expression to be zero is if (2 + 4x = 0). Solving for (x), we get (x = -\frac{1}{2}).
  4. Choose test points on either side of (x = -\frac{1}{2}) to determine the concavity intervals.
    • For (x < -\frac{1}{2}), pick (x = -1). (f''(-1) = (2 + 4(-1))e^{-1} = (-2)e^{-1} < 0), so the function is concave down on (-\infty < x < -\frac{1}{2}).
    • For (x > -\frac{1}{2}), pick (x = 0). (f''(0) = (2 + 4(0))e^{0} = 2 > 0), so the function is concave up on (-\frac{1}{2} < x < \infty).

Therefore, the function (f(x) = x^2e^x) is concave down on (-\infty < x < -\frac{1}{2}) and concave up on (-\frac{1}{2} < x < \infty).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7