Math Problem Statement

Find all the critical numbers of f left parenthesis x right parenthesis equals cube root of 2 x squared minus 18 end root, then determine the local minimum and maximum points by using a graph.

Solution

To find the critical numbers of the function f(x)=2x2183f(x) = \sqrt[3]{2x^2 - 18}, we need to follow these steps:

  1. Find the first derivative of f(x)f(x).
  2. Set the first derivative equal to zero and solve for xx to find critical points.
  3. Check where the first derivative is undefined.
  4. Determine the local minimum and maximum points using a graph.

Step 1: Find the First Derivative

The function is f(x)=(2x218)1/3f(x) = (2x^2 - 18)^{1/3}.

Using the chain rule for differentiation: f(x)=ddx((2x218)1/3)f'(x) = \frac{d}{dx} \left( (2x^2 - 18)^{1/3} \right)

Let u=2x218u = 2x^2 - 18. Then f(x)=u1/3f(x) = u^{1/3}.

Using the chain rule: f(x)=ddu(u1/3)dudxf'(x) = \frac{d}{du} \left( u^{1/3} \right) \cdot \frac{du}{dx} f(x)=13u2/3(4x)f'(x) = \frac{1}{3} u^{-2/3} \cdot (4x) f(x)=4x3(2x218)2/3f'(x) = \frac{4x}{3} (2x^2 - 18)^{-2/3}

Step 2: Set the First Derivative Equal to Zero and Solve for xx

4x3(2x218)2/3=0\frac{4x}{3} (2x^2 - 18)^{-2/3} = 0

This equation is zero when 4x=04x = 0: x=0x = 0

Step 3: Check Where the First Derivative is Undefined

The first derivative is undefined when the denominator is zero: (2x218)2/3=(2x^2 - 18)^{-2/3} = \infty This occurs when: 2x218=02x^2 - 18 = 0 x2=9x^2 = 9 x=±3x = \pm 3

Critical Numbers

The critical numbers are x=0,3,x = 0, 3, and 3-3.

Step 4: Determine Local Minimum and Maximum Points Using a Graph

We need to analyze the behavior of f(x)f(x) at these critical points using the first and second derivative tests, but we can also use a graph to visually determine the local minimum and maximum points.

Plotting f(x)=2x2183f(x) = \sqrt[3]{2x^2 - 18}:

By graphing, we can observe the behavior of the function near the critical points.

Critical Point Analysis

  • At x=0x = 0:

    • f(0)=2(0)2183=183f(0) = \sqrt[3]{2(0)^2 - 18} = \sqrt[3]{-18}
    • The function has an inflection point here since the graph changes concavity but does not achieve a local minimum or maximum.
  • At x=3x = 3:

    • f(3)=2(3)2183=03=0f(3) = \sqrt[3]{2(3)^2 - 18} = \sqrt[3]{0} = 0
    • The function achieves a local minimum here.
  • At x=3x = -3:

    • f(3)=2(3)2183=03=0f(-3) = \sqrt[3]{2(-3)^2 - 18} = \sqrt[3]{0} = 0
    • The function achieves a local maximum here.

Conclusion

  • Critical numbers: x=0,3,3x = 0, 3, -3
  • Local minimum point: (3,0)(3, 0)
  • Local maximum point: (3,0)(-3, 0)

Would you like further details or have any questions?

Here are five questions you might ask:

  1. How do you perform the first and second derivative tests?
  2. Can you explain how to graph a function using critical points?
  3. What is the significance of an inflection point?
  4. How do you determine if a function is increasing or decreasing?
  5. What are some common mistakes to avoid when finding critical points?

Tip: Always check where the derivative is undefined as these points can be critical for identifying the behavior of the function.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
Graphical Analysis

Formulas

Chain Rule
First Derivative Test
Second Derivative Test

Theorems

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Suitable Grade Level

Advanced High School / College