How do i find the extrema (local maximum and minimum) by using the first derivative of: x^3-(3/2)x^2?
Local minimum or maximum is points where first derivation=0
Derivation of x^3-(3/2)x^2 is
3x^2-(3/2)*2x=3x^2-3x
=3*(x^2-x)
=3*x*(x-1)
Zeros of this expresion is x=0 and x=1
Second derivate of this x^3-(3/2)x^2 is
6x-3
Function have local minimum where is second derivation is positive.
Function have local maximum where is second derivation is negative.
Test:
Second derivate=6x-3
x1=0 x2=1
For x=0 6x-3=6*0-3=0-3= -3 Local maximum
For x=1 6x-3=6*1-3=6-3=3 Local minimum
If you wat to see graph of your function in google type:
"function graphs online"
When you see list of results click on:
rechneronline.de/function-graphs
When page be open in blue rectacangle type:
x^3-1.5x^2
Then click option Draw
You will see graph of your function
To find the extrema (local maximum and minimum) of a function using the first derivative, follow these steps:
1. Find the first derivative of the given function. In this case, the function is f(x) = x^3 - (3/2)x^2. To find the derivative, apply the power rule for differentiation:
f'(x) = 3x^2 - 3x.
2. Set the first derivative equal to zero and solve for x. In this case, we have:
3x^2 - 3x = 0.
Factoring out a common factor of 3x, the equation becomes:
3x(x - 1) = 0.
Setting each factor equal to zero, we get two possible solutions:
3x = 0 --> x = 0,
x - 1 = 0 --> x = 1.
3. Determine the critical points. The critical points are the x-values where the first derivative is either zero or undefined. In this case, we have two critical points: x = 0 and x = 1.
4. Test the intervals between the critical points by plugging in values from each interval into the first derivative. We need to determine if the first derivative changes sign in each interval. If the sign changes, there is an extrema in that interval.
For x < 0, choose a test value such as x = -1. Plugging this value into the first derivative, we have:
f'(-1) = 3(-1)^2 - 3(-1) = 3 + 3 = 6.
Since the first derivative is positive (greater than 0), we conclude that there is no extrema in this interval.
For 0 < x < 1, choose a test value such as x = 0.5. Plugging this value into the first derivative, we have:
f'(0.5) = 3(0.5)^2 - 3(0.5) = 0.75 - 1.5 = -0.75.
The first derivative is negative (less than 0) in this interval, indicating a local maximum at x = 0.5.
For x > 1, choose a test value such as x = 2. Plugging this value into the first derivative, we have:
f'(2) = 3(2)^2 - 3(2) = 12 - 6 = 6.
The first derivative is positive (greater than 0) in this interval, indicating a local minimum at x = 2.
5. Summarize the results. The local maximum occurs at x = 0.5, and the local minimum occurs at x = 2.
Therefore, using the first derivative of f(x) = x^3 - (3/2)x^2, we find that the local maximum occurs at x = 0.5, and the local minimum occurs at x = 2.