AboutPaul 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.
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?
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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.)
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