Find relative extrema of f(x)=x^3-3x^2+4
To find relative maxima and minima, first find the critical points (where f�Œ is 0 or doesn�ft exist).
Then examine each critical point.
It is a relative maximum if f�Œ changes from positive to negative or f�� is negative.
It is a relative minimum if f�Œ changes from negative to positive or f�� is positive.
We find the critical numbers of f by solving the equation f'( x ) = 0
f'( x ) = 3 * x ^ 2 - 2 * 3 * x
f'( x ) = 3 x ^ 2 - 6 x
f'( x ) = 3 x ( x - 2 ) = 0
So the critical numbers are x = 0 and x = 2
f�� ( x ) = 3 * 2 * x - 6 = 6 x - 6 = 6 ( x - 1 )
For x = 0
f�� = 6 * ( 0 - 1 ) = 6 * - 1 = - 6
f�� < 0
It is a relative maximum
For x = 2
f�� = 6 * ( 2 - 1 ) = 6 * 1 = 6
It is a relative minimum
Relative maximum :
x = 0
y = 0 ^ 3 - 3 * 0 ^ 2 + 4 = 4
Relative minimum :
x = 2
y = 2 ^ 2 - 3 * 2 ^ 2 + 4 = 8 - 3 * 4 + 4 = 8 - 12 + 4 = 0
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x^3-3x^2+4
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To find relative maxima and minima, first find the critical points (where f´ ( x ) is 0 or doesn´t exist).
Then examine each critical point.
It is a relative maximum if f´ changes from positive to negative or f" is negative.
It is a relative minimum if f´ changes from negative to positive or f" is positive.
f�� mean f "
To find the relative extrema of the function f(x) = x^3 - 3x^2 + 4, we need to find the points where the derivative of the function is equal to zero.
Step 1: Find the derivative f'(x) of the function f(x).
f'(x) = 3x^2 - 6x
Step 2: Set the derivative equal to zero and solve for x.
3x^2 - 6x = 0
Factor out x from both terms:
x(3x - 6) = 0
Then solve for x:
x = 0 or 3x - 6 = 0
If 3x - 6 = 0, then x = 2.
So, the critical points are x = 0 and x = 2.
Step 3: Determine the nature of the critical points using the second derivative test.
To do that, find the second derivative f''(x) of the function f(x):
f''(x) = 6x - 6
Evaluate f''(x) at the critical points x = 0 and x = 2.
At x = 0:
f''(0) = 6(0) - 6 = -6
Since the second derivative is negative at x = 0, it means the function has a local maximum at x = 0.
At x = 2:
f''(2) = 6(2) - 6 = 6
Since the second derivative is positive at x = 2, it means the function has a local minimum at x = 2.
Step 4: Determine the y-values corresponding to the critical points.
To find the y-values, substitute the critical points back into the original function f(x).
For x = 0:
f(0) = 0^3 - 3(0^2) + 4 = 4
For x = 2:
f(2) = 2^3 - 3(2^2) + 4 = 2
So, the relative extrema of the function f(x) = x^3 - 3x^2 + 4 are:
- Local maximum at (0, 4)
- Local minimum at (2, 2)
To find the relative extrema of a function, we will need to find the critical points of the function, where the derivative is either zero or does not exist. We can do this by taking the derivative of the given function and setting it equal to zero.
First, let's find the derivative of f(x) = x^3 - 3x^2 + 4:
f'(x) = 3x^2 - 6x
Now, set the derivative equal to zero and solve for x:
3x^2 - 6x = 0
3x(x - 2) = 0
From here, we have two possibilities:
1. 3x = 0, which gives x = 0.
2. (x - 2) = 0, which gives x = 2.
Now we have found the critical points of the function, which are x = 0 and x = 2.
To determine whether these critical points are relative extrema, we can use the second derivative test or analyze the behavior of the function around these points.
Let's find the second derivative of f(x):
f''(x) = 6x - 6
Now, substitute the critical points into the second derivative:
For x = 0:
f''(0) = 6(0) - 6 = -6, which is a negative number.
For x = 2:
f''(2) = 6(2) - 6 = 6, which is a positive number.
According to the second derivative test, if the second derivative at a critical point is positive, then it is a relative minimum, and if the second derivative is negative, it is a relative maximum.
So, based on the second derivative test, we can conclude:
- At x = 0, f(x) has a relative maximum.
- At x = 2, f(x) has a relative minimum.
Therefore, the relative extrema of f(x) = x^3 - 3x^2 + 4 are:
- Relative maximum at (0, 4)
- Relative minimum at (2, -4)