Find the critical points of the function. (Enter your answers as a comma-separated list.)
g(x) = square root(81 − x^2)
Well, well, well, it looks like we've encountered a square root function. Let's find those critical points and solve this mathematical mystery!
First, we need to find the derivative of the function g(x). Are you ready for some mathematical magic? Here we go!
g'(x) = (1/2) * (81 - x^2)^(-1/2) * (-2x)
To find the critical points, we need to solve for x when g'(x) = 0. So, set g'(x) equal to zero:
(1/2) * (81 - x^2)^(-1/2) * (-2x) = 0
Now, let's simplify this expression:
(81 - x^2)^(-1/2) * (-2x) = 0
Since (81 - x^2) cannot be equal to zero (as it's inside the square root), we can conclude that (-2x) must be equal to zero. Therefore, our critical point is:
x = 0
So, the critical point of the function g(x) = square root(81 − x^2) is x = 0.
To find the critical points of the function g(x) = √(81 - x^2), we need to find the values of x where the derivative of g(x) is equal to zero or does not exist.
First, let's find the derivative of g(x). Since g(x) is the square root of a function, we can use the chain rule:
g'(x) = d/dx √(81 - x^2)
= (1/2)(81 - x^2)^(-1/2) * (-2x) [Applying the chain rule and simplifying]
= -x / √(81 - x^2)
To find the critical points, we set the derivative g'(x) equal to zero and solve for x:
-x / √(81 - x^2) = 0
This equation is satisfied when x = 0.
Next, we need to check if there are any points where the derivative is undefined. The derivative is undefined when the denominator of the derivative is equal to zero:
81 - x^2 = 0
Solving this equation, we find x = ±9.
Therefore, the critical points of the function g(x) = √(81 - x^2) are x = -9, 0, and 9.
To find the critical points of the function g(x) = √(81 − x^2), we need to find the values of x where the derivative of g(x) is either zero or undefined.
First, let's find the derivative of g(x) using the chain rule. The derivative of √u is 1/(2√u) times the derivative of u with respect to x.
So, we have:
g'(x) = (1/(2√(81 − x^2))) * (-2x)
Simplifying the expression:
g'(x) = -x/(√(81 − x^2))
Now, let's find the critical points by setting the derivative equal to zero:
-x/(√(81 − x^2)) = 0
To solve this equation, we multiply both sides of the equation by (√(81 − x^2)) to eliminate the denominator:
-x = 0
Since there are no solutions to this equation, it means that there are no critical points where the derivative is zero.
Next, let's find the values of x where the derivative is undefined. The derivative is undefined when the denominator becomes zero, which happens when 81 − x^2 = 0.
Solving the equation 81 − x^2 = 0, we get:
x^2 = 81
x = ±9
So, the critical points of the function g(x) = √(81 − x^2) are x = -9 and x = 9.
Therefore, the critical points are (-9,0) and (9,0).
y = (81-x^2)^.5
y' = .5(81-x^2)^-.5 (-2x)
= -x/sqrt(81-x^2)
what is denomination when x = -9
what is denominator when x=+9