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.
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.
To solve this problem and find the relative extrema, interval of increasing and decreasing, inflection points, and concavity, we need to go through several steps. Let's take it one step at a time:
A. Finding the relative extrema:
1. Take the first derivative of the function f(x) to find critical points. Critical points occur where the derivative is equal to zero or is undefined.
f'(x) = (1)x^3 - (5/3)x^2 - (3)x + 15
2. Set f'(x) equal to zero and solve for x. In this case, it is a cubic equation, and we can use different methods to solve it like factoring, using the Rational Root Theorem, or using numerical methods like graphing calculators or computer programs.
3. After solving for x, you will obtain the x-values of the critical points. These x-values correspond to the relative extrema of the function f(x). Classify each point as a maximum or minimum by considering the sign changes around the point. If the sign changes from positive to negative, it is a maximum, and if the sign changes from negative to positive, it is a minimum.
B. Determining the interval(s) of increasing and decreasing:
1. Take the first derivative, f'(x), and find the intervals where the derivative is positive and negative.
2. Positive intervals indicate where the function is increasing, and negative intervals indicate where the function is decreasing.
C. Determining the inflection points:
1. Take the second derivative of the function, f''(x), which represents the concavity of the function.
2. Set f''(x) equal to zero and solve for x to find potential inflection points.
3. Similar to step 3 in part A, classify the inflection points using the sign changes of f''(x). If the sign changes from positive to negative or negative to positive, it is an inflection point.
D. Determining concavity and sketching the function:
1. Evaluate the second derivative, f''(x), at the provided points x = -2, -1, 1, and 4.
2. If f''(x) is positive at a point, the function is concave up at that point. If f''(x) is negative, the function is concave down.
3. Use the information from parts A, B, and C to sketch the function. Include the relative extrema, interval of increasing or decreasing, and inflection points.
It's important to note that the calculations involved might be time-consuming and require additional mathematical skills. Using graphing calculators or software can expedite the process.