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.
#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.
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.
You're welcome! It looks like you did a great job with these problems. Let's go through each question and explain how you arrived at your answers.
1. Find all points of inflection: f(x) = (1/12)x^4 - 2x^2 + 15
To find the points of inflection, you need to find the x-values where the concavity of the function changes. Here's how you did it:
- First, you took the second derivative of the function f(x), which is f''(x).
- Then, you set the second derivative equal to zero and solved for x.
- By plugging the obtained x-values back into the original function, you found the corresponding y-values for the points of inflection.
Your answer choice D, (2, 25/3) and (-2, 25/3), is correct. These are the x and y-values of the two points of inflection.
2. Find the absolute maximum and absolute minimum of f on the interval (0,3]: f(x) = (x^3 - 4x^2 + 7x)/x
To find the absolute maximum and absolute minimum on a given interval, you need to find the critical points and compare the function values at these points. Here's how you did it:
- First, you found the first derivative of the function f(x), which is f'(x).
- Then, you identified the critical points by setting the first derivative equal to zero and solving for x.
- After that, you used the interval test to determine if each interval between the critical points is increasing or decreasing.
- Finally, you used the first derivative test to identify the local maximum and local minimum points on the interval (0, 3].
Your answer choice C, Relative maximum: x=2, is correct. The point (2, 3) is the relative maximum on the given interval.
3. Find the values of x that give relative extrema for the function f(x) = (x+1)^2(x-2)
To find the relative extrema, you again need to find the critical points and analyze the intervals using the first derivative test. Here's how you did it:
- First, you found the first derivative of the function f(x), which is f'(x).
- Then, you identified the critical points by setting the first derivative equal to zero and solving for x.
- Next, you used the interval test to determine if each interval between the critical points is increasing or decreasing.
- Finally, you used the first derivative test to identify the relative maximum and minimum points.
Your answer choice A, Relative maximum: x=-1; Relative minimum: x=1, is correct. The point (-1, 0) is the relative maximum, and the point (1, 0) is the relative minimum.
Well done on solving these problems! Your explanations were clear and accurate. Keep up the good work!