find the minimum and maximum points of the curve y=x^4-4x^3+4x^2
min/max occur where y'=0
So, just solve
4x^3-12x^2+8x=0
So, which are min and which max?
Minimum if y" > 0 (concave up)
Maximum if y" < 0 (concave down)
So, take a look at y" = 4(3x^2-6x+2)
To find the minimum and maximum points of the curve y = x^4 - 4x^3 + 4x^2, we can use calculus.
1. First, find the derivative of the function.
The derivative of y = x^4 - 4x^3 + 4x^2 is:
y' = 4x^3 - 12x^2 + 8x
2. Set the derivative equal to zero and solve for x to find the critical points.
4x^3 - 12x^2 + 8x = 0
Factor out an x:
x(4x^2 - 12x + 8) = 0
Simplify the equation:
x(x^2 - 3x + 2) = 0
x(x - 1)(x - 2) = 0
So the critical points are x = 0, x = 1, and x = 2.
3. To determine if each critical point is a minimum or maximum, we need to analyze the second derivative of the function.
Taking the derivative of our derivative, y' = 4x^3 - 12x^2 + 8x, gives us the second derivative:
y'' = 12x^2 - 24x + 8
4. Substitute the critical points (x = 0, 1, and 2) into the second derivative equation and evaluate the value.
For x = 0:
y'' = 12(0)^2 - 24(0) + 8 = 8
Since the second derivative is positive (greater than zero), this means x = 0 is a minimum point.
For x = 1:
y'' = 12(1)^2 - 24(1) + 8 = -4
Since the second derivative is negative (less than zero), this means x = 1 is a maximum point.
For x = 2:
y'' = 12(2)^2 - 24(2) + 8 = 8
Since the second derivative is positive (greater than zero), this means x = 2 is a minimum point.
5. Finally, substitute each critical point (x = 0, 1, and 2) back into the original function to find the corresponding y-values.
For x = 0:
y = (0)^4 - 4(0)^3 + 4(0)^2 = 0
So the minimum point is (0, 0).
For x = 1:
y = (1)^4 - 4(1)^3 + 4(1)^2 = 1 - 4 + 4 = 1
So the maximum point is (1, 1).
For x = 2:
y = (2)^4 - 4(2)^3 + 4(2)^2 = 16 - 32 + 16 = 0
So the minimum point is (2, 0).
Therefore, the minimum points are (0, 0) and (2, 0), and the maximum point is (1, 1).
To find the minimum and maximum points of the curve, we need to find the critical points and determine their nature.
Step 1: Find the derivative of the function.
The derivative of the function y = x^4 - 4x^3 + 4x^2 can be found by applying the power rule. The power rule states that if y = x^n, then the derivative dy/dx is given by dy/dx = nx^(n-1).
Taking the derivative of y = x^4 - 4x^3 + 4x^2:
dy/dx = 4x^3 - 12x^2 + 8x
Step 2: Find the critical points.
To find the critical points, set the derivative equal to zero and solve for x:
4x^3 - 12x^2 + 8x = 0
Step 3: Solve for x.
Factor out the common factor of 4x from the equation:
4x(x^2 - 3x + 2) = 0
Setting each factor equal to zero gives three possible values for x:
1. 4x = 0 -> x = 0
2. x^2 - 3x + 2 = 0
To solve the quadratic equation x^2 - 3x + 2 = 0, we can factor it as (x - 1)(x - 2) = 0, which gives two additional critical points:
3. x - 1 = 0 -> x = 1
4. x - 2 = 0 -> x = 2
So we have three critical points: x = 0, 1, and 2.
Step 4: Determine the nature of the critical points.
To determine the nature of each critical point, we can use the second derivative test. The second derivative represents the concavity of the function and helps us identify whether a critical point corresponds to a minimum, maximum, or neither.
To find the second derivative, differentiate the derivative obtained in Step 1:
d^2y/dx^2 = 12x^2 - 24x + 8
Evaluate the second derivative at each critical point:
1. When x = 0: d^2y/dx^2 = 8 > 0 => This critical point corresponds to a minimum.
2. When x = 1: d^2y/dx^2 = -4 < 0 => This critical point corresponds to a maximum.
3. When x = 2: d^2y/dx^2 = 24 > 0 => This critical point corresponds to a minimum.
Step 5: Find the corresponding y-values.
Substitute the critical points into the original function to find their corresponding y-values:
1. When x = 0: y = 0^4 - 4(0)^3 + 4(0)^2 = 0
2. When x = 1: y = 1^4 - 4(1)^3 + 4(1)^2 = 1 - 4 + 4 = 1
3. When x = 2: y = 2^4 - 4(2)^3 + 4(2)^2 = 16 - 32 + 16 = 0
Therefore, the minimum point is (2, 0) and the maximum point is (1, 1).