Calculus/Graph sketching.
Expert: Paul Klarreich - 9/17/2009
QuestionI was given an assignment and would really appreciate if you can help me with this question.
Find the critical points of function f(x) = x^3 [e^-6x]
Determine whether it is a maximum, minimum or inflexion point
Sketch the graph of y=f(x), indicating the critical points and the values where the graph cuts both the x-axis and y-axis.
thank you~
Answer
Questioner: Stella
Country: Singapore
Category: Calculus
Private: No
Subject: Finding the critical point
Question: I was given an assignment and would really appreciate if you can help me with this question.
Find the critical points of function f(x) = x^3 [e^-6x]
Determine whether it is a maximum, minimum or inflexion point
Sketch the graph of y=f(x), indicating the critical points and the values where the graph cuts both the x-axis and y-axis.
thank you~
..............................................
Find the critical points of function f(x) = x^3 [e^-6x]
The key here is to remember that e^(anything) is always positive, never zero.
You will find f'(x). (product rule, anyone?)
f'(x) = ()() + ()()
f'(x) = (x^3)(-6e^-6x) + (3x^2)(e^-6x)
f'(x) = e^-6x[-6x^3 + 3x^2]
f'(x) = -3x^2e^-6x[2x - 1]
set that = 0, solve, you get
x = 0 and x = 1/2
Now you can do either:
A. The second derivative test: Find f''(x), evaluate it at x = 0, x = 1/2.
When f'' is
-- Positive, you have a min.
-- Negative, you have a max.
-- Zero, you might not be sure.
B. The first derivative test. Use the intervals:
(-inf,0)
(0,1/2)
(1/2, inf)
Pick a value of x in each and find f'(x) for it.
The key here is to remember that e^(anything) is always positive, never zero.
(-inf,0): Pick x = -1: f(-1) = -3[-3 - 1] is positive. [RISING]
(0,1/2): Pick x = 1/4: f(1/4) = -3/4[1/2 - 1] is positive. [RISING]
(1/2, inf): Pick x = 1: f(1) = -3(2 - 1) is negative. [FALLING]
I think x = 0 is an inflection point, x = 1/2 is a max.
I will leave the other stuff to you:
cuts x-axis: Solve f(x) = 0.
cuts y-axis: Find y = f(0)