Math Problem Statement

Find the critical numbers for the given functions: a) f(x) = x^3 - 6x^2 + 9x + 10, b) f(x) = e^x / (x - 2), c) f(x) = 2√(x(6 - x))

Solution

The problem asks to find the critical numbers for the given functions:

Part a: f(x)=x36x2+9x+10f(x) = x^3 - 6x^2 + 9x + 10

  1. Find the derivative of f(x)f(x): f(x)=3x212x+9f'(x) = 3x^2 - 12x + 9
  2. Set the derivative equal to zero to find the critical points: 3x212x+9=03x^2 - 12x + 9 = 0 Divide the equation by 3: x24x+3=0x^2 - 4x + 3 = 0 Factor the quadratic: (x3)(x1)=0(x - 3)(x - 1) = 0 So, the critical points are x=3x = 3 and x=1x = 1.

Part b: f(x)=exx2f(x) = \frac{e^x}{x - 2}

  1. Find the derivative using the quotient rule: f(x)=(x2)(ex)(ex)(1)(x2)2=ex((x2)1)(x2)2=ex(x3)(x2)2f'(x) = \frac{(x - 2)(e^x) - (e^x)(1)}{(x - 2)^2} = \frac{e^x((x - 2) - 1)}{(x - 2)^2} = \frac{e^x(x - 3)}{(x - 2)^2}
  2. Set the derivative equal to zero to find the critical points: ex(x3)(x2)2=0\frac{e^x(x - 3)}{(x - 2)^2} = 0 Since ex0e^x \neq 0, the numerator (x3)=0(x - 3) = 0, so: x=3x = 3 However, we must also check the denominator for restrictions. x=2x = 2 makes the denominator zero, so it's a vertical asymptote and not a critical point. Therefore, the only critical number is x=3x = 3.

Part c: f(x)=2x(6x)f(x) = 2\sqrt{x(6 - x)}

  1. Rewrite the function in a more manageable form: f(x)=26xx2f(x) = 2\sqrt{6x - x^2}
  2. Find the derivative using the chain rule: f(x)=ddx(2(6xx2)1/2)=212(6xx2)1/2(62x)f'(x) = \frac{d}{dx}\left(2(6x - x^2)^{1/2}\right) = 2 \cdot \frac{1}{2}(6x - x^2)^{-1/2} \cdot (6 - 2x) Simplifying the derivative: f(x)=62x6xx2f'(x) = \frac{6 - 2x}{\sqrt{6x - x^2}}
  3. Set the derivative equal to zero: 62x6xx2=0\frac{6 - 2x}{\sqrt{6x - x^2}} = 0 So, 62x=06 - 2x = 0, which gives: x=3x = 3
  4. Check the domain: The function is defined where 6xx206x - x^2 \geq 0, or x(6x)0x(6 - x) \geq 0, which implies 0x60 \leq x \leq 6.

Thus, the critical point is x=3x = 3.


Would you like more details or have any questions?

Here are 5 related questions for practice:

  1. What are the critical points for f(x)=x44x3+6x2f(x) = x^4 - 4x^3 + 6x^2?
  2. How do you apply the quotient rule to a more complex function like sinxx2\frac{\sin x}{x^2}?
  3. Can a critical point also be an inflection point?
  4. How do you determine if a critical point is a minimum, maximum, or neither?
  5. How can you apply implicit differentiation to find critical points of implicit functions?

Tip: Critical numbers occur where the derivative is zero or undefined, but make sure to check the domain of the function!

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Critical Points
Quotient Rule
Chain Rule

Formulas

f'(x) = 3x^2 - 12x + 9 for function a
Quotient rule: f'(x) = [(x - 2)(e^x) - (e^x)(1)] / (x - 2)^2 for function b
f'(x) = (6 - 2x) / √(6x - x^2) for function c

Theorems

Critical points occur where f'(x) = 0 or f'(x) is undefined
Quotient Rule
Chain Rule

Suitable Grade Level

Grades 10-12