Math Problem Statement

The stationary points of y = x^3 + 4x^2 - 5x + 2 are (__, __).

Solution

To find the stationary points of the given function y=x3+4x25x+2y = x^3 + 4x^2 - 5x + 2, we need to follow these steps:

  1. Find the first derivative of the function. Stationary points occur where the first derivative equals zero.

dydx=3x2+8x5\frac{dy}{dx} = 3x^2 + 8x - 5

  1. Set the derivative equal to zero to find the x-coordinates of the stationary points:

3x2+8x5=03x^2 + 8x - 5 = 0

This is a quadratic equation. Solving it will give us the x-coordinates of the stationary points. Let's solve this equation now.

I'll calculate the solutions.The solutions for xx are:

x1=4+313,x2=4313x_1 = \frac{-4 + \sqrt{31}}{3}, \quad x_2 = \frac{-4 - \sqrt{31}}{3}

Now, let's calculate the approximate decimal values for these solutions to two decimal places.The stationary points occur at x1=0.52x_1 = 0.52 and x2=3.19x_2 = -3.19.

Would you like further details or explanations? Here are some related questions:

  1. How do you classify stationary points (maximum, minimum, or inflection)?
  2. What is the second derivative test for stationary points?
  3. Can stationary points occur at any place other than where the derivative is zero?
  4. What is the difference between stationary points and turning points?
  5. How do you graph a function based on its stationary points?

Tip: After finding the stationary points, use the second derivative test to determine if they are maxima, minima, or points of inflection.

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Math Problem Analysis

Mathematical Concepts

Calculus
Stationary Points
Derivatives

Formulas

Derivative of y with respect to x: dy/dx = 3x^2 + 8x - 5
Quadratic equation: ax^2 + bx + c = 0
Quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / 2a

Theorems

Stationary points occur where the first derivative equals zero
Quadratic formula for solving second-degree equations

Suitable Grade Level

Grades 10-12