How do you find the critical points and the open intervals where the function is increasing and decreasing for #y = xe^(x(2 - 3x))#?

Answer 1

The function is increasing on the interval: #((1-sqrt7)/6, (1+sqrt7)/6)# and decreasing on the intervals: #(-oo, (1-sqrt7)/6)# and #((1+sqrt7)/6, oo)#

Jasivan S, gives an excellent answer to the question asked, but I suspect that the intended question was: Find the open intervals on which the function is increasing and those on which it is decreasing. (Rather than finding those on which it is doing both.)

See Jasivan S. answer to find the derivative: #dy/dx=(-6x^2+2x+1)e^(2x-3x^2)#
and the critical numbers: #x_1=(1+sqrt(7))/6# and #x_2=(1-sqrt(7))/6#.
(I'm used to thinking of critical points as points in the domain, so these would be the critical points. Some think of critical points as points on a graph, in which case, you'd need the #y# values as well. See Jasivan S. answer for that.)
As you probably know, to investigate increasing and decreasing behavior, we consider the sign of #dy/dx#.
Because #e^23-3x^2)# is always positive, the sign of #dy/dx# will be the same as the sign of #-6x^2+2x+1# which is positive on #((1-sqrt7)/6, (1+sqrt7)/6)#, and negative on #(-oo, (1-sqrt7)/6)# and on #((1+sqrt7)/6, oo)#.

Cosequently,

The function is increasing on the interval: #((1-sqrt7)/6, (1+sqrt7)/6)# and decreasing on the intervals: #(-oo, (1-sqrt7)/6)# and #((1+sqrt7)/6, oo)#
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Answer 2

To find the critical points and the intervals where the function is increasing or decreasing for ( y = xe^{x(2 - 3x)} ), follow these steps:

  1. Find the derivative of the function ( y ) with respect to ( x ), denoted as ( y' ).
  2. Set ( y' ) equal to zero and solve for ( x ) to find the critical points.
  3. Determine the intervals where ( y' > 0 ) to identify where the function is increasing.
  4. Determine the intervals where ( y' < 0 ) to identify where the function is decreasing.

Let's go through these steps:

  1. The derivative of the function ( y = xe^{x(2 - 3x)} ) can be found using the product rule and chain rule:

[ y' = e^{x(2 - 3x)} + x(e^{x(2 - 3x)})\cdot(2 - 3x) ]

  1. Set ( y' ) equal to zero and solve for ( x ) to find the critical points:

[ e^{x(2 - 3x)} + x(e^{x(2 - 3x)})\cdot(2 - 3x) = 0 ]

This equation may not have an explicit solution, so you may need to use numerical methods or graphing techniques to approximate the critical points.

  1. Determine the intervals where ( y' > 0 ) to identify where the function is increasing.

  2. Determine the intervals where ( y' < 0 ) to identify where the function is decreasing.

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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.

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