Math Problem Statement

f(x)=-1x^3+5x^2-7x-6

Solution

To analyze the function f(x)=x3+5x27x6f(x) = -x^3 + 5x^2 - 7x - 6, we can explore various characteristics such as its intercepts, critical points, and end behavior. Here’s a breakdown of each:

1. Y-intercept

To find the y-intercept, substitute x=0x = 0: f(0)=03+502706=6f(0) = -0^3 + 5 \cdot 0^2 - 7 \cdot 0 - 6 = -6 So, the y-intercept is at (0,6)(0, -6).

2. X-intercepts

Finding the x-intercepts involves solving f(x)=0f(x) = 0: x3+5x27x6=0-x^3 + 5x^2 - 7x - 6 = 0 This cubic equation may have real roots. You could attempt to factor it, use synthetic division, or apply numerical methods to find approximate roots.

3. First Derivative (Critical Points)

To find the critical points, we calculate the first derivative and set it to zero: f(x)=3x2+10x7f'(x) = -3x^2 + 10x - 7 Solving f(x)=0f'(x) = 0: 3x2+10x7=0-3x^2 + 10x - 7 = 0 This is a quadratic equation, which can be solved using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Plugging in a=3a = -3, b=10b = 10, and c=7c = -7, we get: x=10±1004(3)(7)2(3)=10±100846=10±166x = \frac{-10 \pm \sqrt{100 - 4(-3)(-7)}}{2(-3)} = \frac{-10 \pm \sqrt{100 - 84}}{-6} = \frac{-10 \pm \sqrt{16}}{-6} x=10±46x = \frac{-10 \pm 4}{-6} This gives two solutions: x=10+46=1andx=1046=732.33x = \frac{-10 + 4}{-6} = 1 \quad \text{and} \quad x = \frac{-10 - 4}{-6} = \frac{7}{3} \approx 2.33 These are the critical points: x=1x = 1 and x2.33x \approx 2.33.

4. Second Derivative (Concavity)

The second derivative will help us determine concavity and any possible inflection points: f(x)=6x+10f''(x) = -6x + 10 Setting f(x)=0f''(x) = 0: 6x+10=0-6x + 10 = 0 Solving for xx: x=106=531.67x = \frac{10}{6} = \frac{5}{3} \approx 1.67 At x1.67x \approx 1.67, the concavity may change, suggesting an inflection point at this x-value.

5. End Behavior

Since the leading term of f(x)f(x) is x3-x^3, as xx \to \infty, f(x)f(x) \to -\infty, and as xx \to -\infty, f(x)f(x) \to \infty. This tells us the function has opposing end behaviors due to the odd degree and negative leading coefficient.

Would you like further details on any of these steps or have specific questions?


Related Questions:

  1. How would you solve for exact x-intercepts if factoring is not feasible?
  2. Can we determine the nature of the critical points (maximum or minimum)?
  3. What is the significance of the inflection point in the graph's shape?
  4. How does the function’s end behavior affect the overall graph?
  5. How can we approximate the x-intercepts using numerical methods?

Tip:

When analyzing polynomials, always check the highest-degree term's coefficient for clues on end behavior!

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Math Problem Analysis

Mathematical Concepts

Polynomials
Cubic Functions
Critical Points
End Behavior
Derivative Calculations

Formulas

First Derivative: f'(x) = -3x^2 + 10x - 7
Second Derivative: f''(x) = -6x + 10
Quadratic Formula: x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Critical Points Theorem
End Behavior of Polynomials

Suitable Grade Level

Grades 11-12