Math Problem Statement

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) F(x) = x4/5(x − 1)2

Solution

To find the critical numbers of the function F(x)=x4/5(x1)2F(x) = x^{4/5}(x - 1)^2, we will follow these steps:

Step 1: Find the derivative of the function

We apply the product rule to differentiate F(x)=x4/5(x1)2F(x) = x^{4/5}(x - 1)^2.

Let:

  • u(x)=x4/5u(x) = x^{4/5}
  • v(x)=(x1)2v(x) = (x - 1)^2

The product rule states: F(x)=u(x)v(x)+u(x)v(x)F'(x) = u'(x)v(x) + u(x)v'(x)

Now we calculate the derivatives of u(x)u(x) and v(x)v(x):

  • u(x)=45x1/5u'(x) = \frac{4}{5}x^{-1/5}
  • v(x)=2(x1)v'(x) = 2(x - 1)

Using the product rule: F(x)=(45x1/5)(x1)2+x4/5(2(x1))F'(x) = \left(\frac{4}{5}x^{-1/5}\right)(x - 1)^2 + x^{4/5}(2(x - 1))

Step 2: Simplify the derivative

Now, simplify the expression for F(x)F'(x):

F(x)=45x1/5(x1)2+2x4/5(x1)F'(x) = \frac{4}{5}x^{-1/5}(x - 1)^2 + 2x^{4/5}(x - 1)

Factor out x1/5x^{-1/5} and (x1)(x - 1) from both terms: F(x)=x1/5(x1)(45(x1)+2x)F'(x) = x^{-1/5}(x - 1)\left( \frac{4}{5}(x - 1) + 2x \right)

Simplify the terms inside the parentheses:

= x^{-1/5}(x - 1)\left( \frac{4}{5}x - \frac{4}{5} + 2x \right)$$ Combine like terms: $$F'(x) = x^{-1/5}(x - 1)\left( \frac{14}{5}x - \frac{4}{5} \right)$$ ### Step 3: Find the critical points To find the critical points, we set $$F'(x) = 0$$. This will occur when any of the factors in $$F'(x) = x^{-1/5}(x - 1)\left( \frac{14}{5}x - \frac{4}{5} \right)$$ is equal to zero: 1. $$x^{-1/5} = 0$$ (this does not give any solution since no power of $$x$$ can be zero). 2. $$x - 1 = 0$$ implies $$x = 1$$. 3. $$\frac{14}{5}x - \frac{4}{5} = 0$$ implies: $$\frac{14}{5}x = \frac{4}{5}$$ $$14x = 4$$ $$x = \frac{4}{14} = \frac{2}{7}$$ ### Step 4: Conclusion The critical numbers are $$x = 1$$ and $$x = \frac{2}{7}$$. Thus, the critical numbers are: $$\boxed{1, \frac{2}{7}}$$ Would you like further details or have any questions? --- Here are 5 related questions to explore this topic further: 1. How do you find critical numbers for a function involving both products and powers? 2. What role do derivatives play in identifying critical points? 3. Can critical numbers exist if a function is not differentiable at certain points? 4. How does the product rule help in differentiating more complex functions? 5. What is the significance of critical numbers in understanding the graph of a function? **Tip:** Critical points occur where the derivative is zero or undefined. Make sure to check for both when solving.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points

Formulas

Product rule: (u*v)' = u'v + uv'
Power rule: d/dx[x^n] = nx^(n-1)

Theorems

Critical point theorem: A function's critical points occur where the derivative is zero or undefined.

Suitable Grade Level

Undergraduate Calculus