#1 there are 4 points of inflection, so (E)
#2 f = x^2-4x+7
f has a min at x=2, so C or D
f(0) -> 7
f(3) = 4
So, (d)
#3 is correct
On stuff like this, it is always helpful to verify your calculations with a graphing utility, of which there are many online.
1. Find all points of inflection: f(x)=1/12x^4-2x^2+15
A. (2, 0)
B. (2, 0), (-2, 0)
C. (0, 15)
D. (2, 25/3), (-2, 25/3)
E. none of these
I got D. I found the second derivative and equaled it to 0 and solved for x. I plugged the x values in to get my points.
2. Find the absolute maximum and absolute minimum of f on the interval (0,3]: f(x)=(x^3-4x^2+7x)/x
A. Maximum: None; Minimum (3, 4)
B. Maximum: (0, 7); Minimum (3, 4)
C. Maximum: None; Minimum (2, 3)
D. Maximum (0, 7); Minimum (2, 3)
E. None of these
I got C. I found the first derivative and critical numbers. I used the interval test to determine increasing or decreasing then I used the first derivative test to determine where the max or min was.
3. Find the values of x that give relative extrema for the function f(x)=(x+1)^2(x-2)
A. Relative maximum: x=-1; Relative minimum: x=1
B. Relative maxima: x=1, x=3; Relative minimum: x=-1
C. Relative maximum: x=2
D. Relative maximum: x=-1; Relative minimum: x=2
E. None of these
I got A. I found the first derivative and critical numbers. I used the interval test to determine increasing or decreasing then I used the first derivative test to determine where the max or min was.
Thank you for checking my answers.
3 answers
When I solve 1 without a graphing utility I one get 2 inflection points. How do you get the other two without a graphing utility. Not allowed to use calculators for this.
That's a nasty one, all right.
f"(x) = 12(240x^6-36x^4-178x^2+5)/(12x^4-2x^2+15)^3
so the numerator is a cubic in x^2, not easy to solve. Not sure why they threw this curve ball.
f"(x) = 12(240x^6-36x^4-178x^2+5)/(12x^4-2x^2+15)^3
so the numerator is a cubic in x^2, not easy to solve. Not sure why they threw this curve ball.