Asked by Steph
Consider the function f(x) = 1/4x^4 - 5/3x^3 - 3/2x^2 + 15x - 3.
A. Find the relative extrema for f(x); be sure to label each as a maximum or minimum. You do not need to find function values; just find the x-values.
B. Determine the interval(s) where f(x) is increasing (if any) and the interval(s) where f(x) is decreasing (if any).
C. Determine the number of inflection point(s) the function has and their x-coordinate(s). Justify your work.
D. Determine whether f(x) is concave up or down at the following points: x = -2, x = -1, x = 1, and x = 4. Use this information and the information in parts A-C to sketch this function.
Please help! I don't understand any of this.
A. Find the relative extrema for f(x); be sure to label each as a maximum or minimum. You do not need to find function values; just find the x-values.
B. Determine the interval(s) where f(x) is increasing (if any) and the interval(s) where f(x) is decreasing (if any).
C. Determine the number of inflection point(s) the function has and their x-coordinate(s). Justify your work.
D. Determine whether f(x) is concave up or down at the following points: x = -2, x = -1, x = 1, and x = 4. Use this information and the information in parts A-C to sketch this function.
Please help! I don't understand any of this.
Answers
Answered by
Reiny
When you say, "I don't understand any of this", I am troubled. This question is as fundamental in beginning Calculus as it gets. Are you taking this course in a traditional classroom setting?
What is your background in math that enables you to take Calculus?
anyway....
f(x) = y = 1/4x^4 - 5/3x^3 - 3/2x^2 + 15x - 3
dy/dx = x^3 - 5x^2 - 3x + 15
a) for extrema, or max/mins, dy/dx = 0
x^3 - 5x^2 - 3x + 15 = 0
looks like grouping will work nicely here
x^2(x - 5) - 3(x - 5) = 0
(x-5)(x^2 - 3) = 0
(x-5)(x - √3)(x + √3) = 0
x = 5 or x = √3 or x = -√3
those are the x's that will produce your max/mins
At this point I will stop to see if you are going to look at this reply, since you posted it last night. Respond and I will continue with the reply.
What is your background in math that enables you to take Calculus?
anyway....
f(x) = y = 1/4x^4 - 5/3x^3 - 3/2x^2 + 15x - 3
dy/dx = x^3 - 5x^2 - 3x + 15
a) for extrema, or max/mins, dy/dx = 0
x^3 - 5x^2 - 3x + 15 = 0
looks like grouping will work nicely here
x^2(x - 5) - 3(x - 5) = 0
(x-5)(x^2 - 3) = 0
(x-5)(x - √3)(x + √3) = 0
x = 5 or x = √3 or x = -√3
those are the x's that will produce your max/mins
At this point I will stop to see if you are going to look at this reply, since you posted it last night. Respond and I will continue with the reply.
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