Math Problem Statement

Graph the function. Identify the point where the local maximum occurs. y = 3x^3 + 7x^2 - 5x - 4

Solution

To identify the point where the local maximum occurs for the function

y=3x3+7x25x4,y = 3x^3 + 7x^2 - 5x - 4,

we need to follow these steps:

  1. Find the First Derivative: Calculate yy' to find the critical points by setting y=0y' = 0.

    y=ddx(3x3+7x25x4)=9x2+14x5.y' = \frac{d}{dx}(3x^3 + 7x^2 - 5x - 4) = 9x^2 + 14x - 5.

  2. Set the Derivative Equal to Zero: Solve 9x2+14x5=09x^2 + 14x - 5 = 0 for xx.

    This is a quadratic equation in xx, which can be solved using the quadratic formula:

    x=b±b24ac2a,x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},

    where a=9a = 9, b=14b = 14, and c=5c = -5.

  3. Calculate the Values of xx:

    x=14±14249(5)29.x = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 9 \cdot (-5)}}{2 \cdot 9}.

    Simplifying inside the square root:

    x=14±196+18018=14±37618.x = \frac{-14 \pm \sqrt{196 + 180}}{18} = \frac{-14 \pm \sqrt{376}}{18}.

    x=14±19.3918.x = \frac{-14 \pm 19.39}{18}.

    This gives two solutions:

    x10.299andx21.855.x_1 \approx 0.299 \quad \text{and} \quad x_2 \approx -1.855.

  4. Determine Which is a Local Maximum: Use the second derivative test or plug these values into the original function to see which gives a maximum.

  5. Evaluate the Function at Each Critical Point: Calculate y(x1)y(x_1) and y(x2)y(x_2), then identify the local maximum.

Let me complete these calculations to confirm the exact values and find the local maximum point.The critical points are approximately:

  • x10.299x_1 \approx 0.299 with y(x1)4.789y(x_1) \approx -4.789
  • x21.855x_2 \approx -1.855 with y(x2)10.213y(x_2) \approx 10.213

Since y(x2)y(x_2) is greater, the local maximum occurs at the point:

(1.855,10.213)(-1.855, 10.213)

Let me know if you'd like further details on this solution or have any questions!

Here are some related questions for further practice:

  1. How do you find local minima using derivatives?
  2. What is the significance of the second derivative in determining concavity?
  3. Can a cubic function have more than one local maximum or minimum?
  4. How would the result change if the coefficients of the function were altered?
  5. What happens to the maximum point if we apply a vertical shift to the function?

Tip: When solving for critical points, always check the function values to distinguish between maximum and minimum points.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivative
Critical Points
Local Maximum

Formulas

First derivative of the function y' = d/dx (3x^3 + 7x^2 - 5x - 4)
Quadratic formula for finding roots: x = (-b ± sqrt(b^2 - 4ac)) / 2a

Theorems

First Derivative Test
Second Derivative Test

Suitable Grade Level

Grades 11-12