Math Problem Statement

Find any local maximum and/or minimum value(s) of the following functions. In each case, draw a sketchy graph of the function. y = x 3 − 9x 2 + 15x + 5

Solution

We are tasked with finding any local maximum and/or minimum values of the function:

y=x39x2+15x+5y = x^3 - 9x^2 + 15x + 5

Steps to find local extrema:

  1. First Derivative (Critical Points): To find the critical points, we first take the derivative of the function and set it equal to zero.

y=ddx(x39x2+15x+5)=3x218x+15y' = \frac{d}{dx} (x^3 - 9x^2 + 15x + 5) = 3x^2 - 18x + 15

Set the derivative equal to zero to find the critical points:

3x218x+15=03x^2 - 18x + 15 = 0

Simplify:

x26x+5=0x^2 - 6x + 5 = 0

Factor the quadratic:

(x5)(x1)=0(x - 5)(x - 1) = 0

So, the critical points are x=5x = 5 and x=1x = 1.

  1. Second Derivative (Concavity Test): To determine if these critical points are maxima or minima, we take the second derivative.

y=ddx(3x218x+15)=6x18y'' = \frac{d}{dx} (3x^2 - 18x + 15) = 6x - 18

Now, evaluate the second derivative at the critical points:

  • At x=1x = 1:

y(1)=6(1)18=12(concave down, local maximum at x=1)y''(1) = 6(1) - 18 = -12 \quad (\text{concave down, local maximum at } x = 1)

  • At x=5x = 5:

y(5)=6(5)18=12(concave up, local minimum at x=5)y''(5) = 6(5) - 18 = 12 \quad (\text{concave up, local minimum at } x = 5)

Local Maximum and Minimum:

  • Local maximum at x=1x = 1. To find the value of yy at x=1x = 1:

y(1)=139(1)2+15(1)+5=19+15+5=12y(1) = 1^3 - 9(1)^2 + 15(1) + 5 = 1 - 9 + 15 + 5 = 12

Thus, the local maximum is at (1,12)(1, 12).

  • Local minimum at x=5x = 5. To find the value of yy at x=5x = 5:

y(5)=539(5)2+15(5)+5=125225+75+5=20y(5) = 5^3 - 9(5)^2 + 15(5) + 5 = 125 - 225 + 75 + 5 = -20

Thus, the local minimum is at (5,20)(5, -20).

Sketch of the Function:

The graph of the function y=x39x2+15x+5y = x^3 - 9x^2 + 15x + 5 has a local maximum at (1,12)(1, 12) and a local minimum at (5,20)(5, -20). The function tends to ++\infty as xx \to \infty and -\infty as xx \to -\infty, since it's a cubic function.

I will now create a rough sketch of this function for you.

Summary:

  • Local maximum at (1,12)(1, 12)
  • Local minimum at (5,20)(5, -20)

Let me know if you would like further clarifications or the image of the graph.


Relative Questions:

  1. What is the significance of the second derivative in determining concavity?
  2. How would a function with a local maximum and minimum differ if it was a quadratic rather than cubic?
  3. What is the behavior of a cubic function as xx \to \infty and xx \to -\infty?
  4. Can there be more than one local maximum or minimum for a cubic function? Why or why not?
  5. How would we find inflection points for a cubic function?

Tip: Always check both the first and second derivatives to confirm the nature of critical points (maximum, minimum, or neither).

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
Concavity
Extrema

Formulas

First derivative: y' = 3x^2 - 18x + 15
Second derivative: y'' = 6x - 18
Quadratic equation: x^2 - 6x + 5 = 0

Theorems

First Derivative Test
Second Derivative Test

Suitable Grade Level

Grades 11-12 / Early College