Find any absolute max/min and local max/min for the function f(x)=x^3-4x+5 on the interval [0,5). Made sure to prove that these points are max/min values.
This is what I did so far.
f'(x)=3x^2-4=0
3x^2=4
x= 2/sqrt3 The -2/sqrt3 is outside of interval so can't use.
x=2/sqrt3
f(0)=5
lim x-->5- is 110. Can't use this though because 5 is not included in the interval.
Since this is the largest of the three numbers, 2/sqrt3, 5, 110, there is no absolute max.
The local minimum is at x=2/sqrt3.
I don't know if this is right or if there are any local max or absolute minimum.
Please help.
To find the absolute max/min and local max/min values of a function on a given interval, you need to follow these steps:
1. Find the critical points of the function within the interval. These are the points where the derivative is either zero or undefined.
To find the critical points of the function f(x) = x^3 - 4x + 5, we need to find where its derivative f'(x) = 3x^2 - 4 is equal to zero:
3x^2 - 4 = 0
3x^2 = 4
x^2 = 4/3
x = ±√(4/3) = ±(2/√3)
Since -2/√3 is outside the given interval [0, 5), we only consider x = 2/√3 as a potential critical point within the interval.
2. Check if the critical points are within the interval.
In this case, x = 2/√3 is within the interval [0, 5).
3. Evaluate the function at the critical points and the endpoints of the interval.
Now, let's substitute the critical point and the endpoints into the original function f(x) = x^3 - 4x + 5:
f(0) = (0)^3 - 4(0) + 5 = 5
f(5) = (5)^3 - 4(5) + 5 = 125 - 20 + 5 = 110
f(2/√3) = (2/√3)^3 - 4(2/√3) + 5 = 17/3 - 8/√3 + 5
Since x = 2/√3 is the only critical point within the interval, we only need to compare f(2/√3) with the function values at the endpoints.
4. Determine the absolute max/min and local max/min values.
From the calculations above, we have:
f(0) = 5
f(2/√3) = 17/3 - 8/√3 + 5
We can see that f(0) = 5 is the largest value among the three numbers: 5, 110, 17/3 - 8/√3 + 5.
Therefore, f(0) = 5 is the absolute maximum value on the interval [0, 5).
Now, to determine if there are any local max/min values, we need to check if f(2/√3) is a local minimum or maximum.
To do this, we can analyze the concavity of the function. Taking the second derivative of f(x):
f''(x) = (3x^2 - 4)' = 6x
Since f''(x) = 6x is positive for all x > 0, we can conclude that f(x) is concave up on the interval [0, 5). Hence, f(2/√3) = 17/3 - 8/√3 + 5 is a local minimum.
So, to summarize:
Absolute maximum: f(0) = 5
Absolute minimum: None within the given interval
Local maximum: None within the given interval
Local minimum: f(2/√3) = 17/3 - 8/√3 + 5
Remember, when determining local max/min values, you should also consider the behavior of the function at the endpoints. In this case, since the interval does not include the endpoint x = 5, it is not considered in the analysis.