Consider the function f(x)=sqrt(2–5x^2), –5≤x≤1. The absolute minimum value is ____ and this occurs at x equals ______?
Found the max value to be 2 at x=0.
f'(x) = (1/2)(2 - 5x^2)^(-1/2) )(-10x)
= -5x(2-5x^2)^(-1/2)
= 0 for a max/min
This has only one solution, when x = 0
so the min occurs when x = 0 and
f(0) = √(2-0) = √2
4x2+24x+16y2-32y-12=0 need to know how to write in the standard form for ellipse it is a precalculus
To find the absolute minimum value and the corresponding x-value of the function f(x) = √(2–5x^2), we need to first find the critical points (where the derivative equals zero or does not exist) and then evaluate the function at these points, including the endpoints of the given interval.
1. Take the derivative of the function to find the critical points:
f'(x) = (-5x) / √(2–5x^2)
2. Set f'(x) equal to zero and solve for x:
(-5x) / √(2–5x^2) = 0
Since the numerator is equal to zero, we have:
-5x = 0
x = 0
This is a critical point.
3. Now, we evaluate the function at x = 0 and the endpoints of the given interval:
f(-5) = √(2–5(-5)^2) = √(2–125) = √(-123) (This is not a real number)
f(1) = √(2–5(1)^2) = √(2–5) = √(-3) (This is not a real number)
f(0) = √(2–5(0)^2) = √2 (This is a real number)
We can see that the function is not defined for x = -5 and x = 1 as it results in imaginary numbers.
Hence, the absolute minimum value of the function does not exist within the given interval.
To find the absolute minimum value and its corresponding value of x for the given function, we need to follow these steps:
1. Find the critical points of the function by taking the derivative and setting it equal to zero.
Let's start by finding the derivative of the function f(x) = √(2 - 5x^2). Using the chain rule, we have:
f'(x) = (1/2)*(2 - 5x^2)^(-1/2) * (-10x)
Simplifying further:
f'(x) = -5x / √(2 - 5x^2)
Now, we set f'(x) equal to zero and solve for x:
-5x / √(2 - 5x^2) = 0
Since a fraction is equal to zero if the numerator is equal to zero, we have:
-5x = 0
x = 0
2. Evaluate the function at the critical points and endpoints of the given interval.
In our case, the interval is -5 ≤ x ≤ 1. So, we need to evaluate the function f(x) = √(2 - 5x^2) at x = -5, 0, and 1.
f(-5) = √(2 - 5(-5)^2) = √(2 - 125) = √(-123) (This is not a real number)
f(0) = √(2 - 5(0)^2) = √(2 - 0) = √(2) = √2
f(1) = √(2 - 5(1)^2) = √(2 - 5) = √(-3) (This is not a real number)
3. Identify the minimum value and its corresponding value of x.
Since f(-5) and f(1) are not real numbers, we do not have minimum values at these points. Therefore, we only need to consider the critical point we found at x = 0.
The function evaluated at x = 0 is f(0) = √2, which is the minimum value on the interval -5 ≤ x ≤ 1.
Therefore, the absolute minimum value is √2, and it occurs at x = 0.