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Calculus/Decreasing Inverval

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Expert: Paul Klarreich - 7/30/2008


Question
Hi,

Here is my question. We're not supposed to use the calculator for this.

On what interval(s) is the function f(x)=x^2e^-x/2 decreasing?

How do I do this problem?

Answer

Intervals
Questioner:   Emma
Category:  Calculus
Private:  No
 
Subject:  Decreasing Inverval
Question:  Hi,

Here is my question. We're not supposed to use the calculator for this.

On what interval(s) is the function f(x)=x^2e^-x/2 decreasing?

How do I do this problem?
.....................................
Hi, Emma,

Try the derivative first:


f(x) = x^2e^(-x/2)  << parenthesize to avoid confusion.

Product rule:

f'(x) = x^2(-1/2  e^(-x/2)) + (2x)(e^(-x/2))

Factor:

f'(x) = [ x^2(-1/2) + 2x ](e^(-x/2))

Now you want to know where that is negative.  But e^(-x/2) is always positive (exponentials usually are).

So you want:  - x^2/2 + 2x  to be negative.  Its graph would be a parabola with its vertex at the top (facing down).  So its negative parts would be OUTSIDE its intercepts.

Solve  - x(x/2 - 2) = 0

Gives  x = 0 and  x = 4.

Your negative intervals are (-inf,0) and (4,+inf).  That's where the graph is falling.

(see attached pic, assuming I remember to attach it.)

Paul Klarreich

Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

Experience

I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

Education/Credentials
(See above.)

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