find the critical numbers of the function
F(x)=x^(4/5)(x-1)^2
To find the critical numbers of the function F(x) = x^(4/5)(x-1)^2, we need to determine the values of x where the derivative of the function is either zero or undefined.
First, let's find the derivative of F(x). Note that we can use the product rule to differentiate the function.
F'(x) = (4/5)x^(-1/5)(x-1)^2 + x^(4/5)(2(x-1))
Next, we will set F'(x) equal to zero and solve for x.
(4/5)x^(-1/5)(x-1)^2 + x^(4/5)(2(x-1)) = 0
To simplify the expression, let's multiply both sides of the equation by 5 to get rid of the fractions.
4x^(-1/5)(x-1)^2 + 5x^(4/5)(2(x-1)) = 0
Now, simplify further:
4(x-1)^2/x^(1/5) + 10(x-1)x^(4/5) = 0
To proceed, let's multiply both sides by x^(1/5) to eliminate the fractional exponents.
4(x-1)^2 + 10(x-1)x = 0
Expanding and collecting like terms:
4(x^2 - 2x + 1) + 10(x^2 - x) = 0
4x^2 - 8x + 4 + 10x^2 - 10x = 0
14x^2 - 18x + 4 = 0
Next, let's factor the quadratic equation:
(7x - 2)(2x - 2) = 0
Setting each factor equal to zero:
7x - 2 = 0... (1)
2x - 2 = 0... (2)
Solving equation (1) for x:
7x = 2
x = 2/7
Solving equation (2) for x:
2x = 2
x = 1
Therefore, the critical numbers of the function F(x) = x^(4/5)(x-1)^2 are x = 2/7 and x = 1.
To find the critical numbers of the function F(x) = x^(4/5)(x-1)^2, we need to find the values of x where the derivative of the function is either zero or undefined.
1. Start by finding the derivative of the function F(x). Let's denote the derivative as F'(x):
F'(x) = (4/5)x^(-1/5)(x-1)^2 + x^(4/5)(2)(x-1)
2. Simplify the derivative by distributing and combining like terms:
F'(x) = (4/5)x^(-1/5)(x-1)^2 + 2x^(4/5)(x-1)
3. To determine where the derivative is undefined, we need to identify any values of x that would make the expression x^(-1/5) undefined. Since any number raised to the power of -1/5 is equivalent to taking the fifth root of that number, we see that the derivative is undefined when x = 0.
4. To find where the derivative is equal to zero, we need to solve the equation F'(x) = 0. Let's set F'(x) = 0 and solve for x:
(4/5)x^(-1/5)(x-1)^2 + 2x^(4/5)(x-1) = 0
We can multiply through by 5 to get rid of the fraction:
4x^(-1/5)(x-1)^2 + 10x^(4/5)(x-1) = 0
5. We can solve this equation using algebraic techniques or approximation methods such as numerical approximation or graphing.
Already factored, nice
So all we have to do is set each factor equal to zero
x^(4/5) = 0 ----> x = 0 , f(0) = 0
(x-1)^2 = 0 ----> x = 1 , f(1) = 0
F ' (x) = x^(4/5) (2(x-1)) + (4/5)x^(-1/5) (x-1)^2
= (2/5)x^(-1/5) (x-1) [5x + 2(x-1)]
= (2/5)(x-1)(7x-2)/x^(1/5
= 0 for max/mins
x = 1 or x = 7/2
A quick sketch in Wolfram will show how these affect the graph
http://www.wolframalpha.com/input/?i=F%28x%29%3Dx%5E%284%2F5%29%28x-1%29%5E2