Find the x-values of all points where the function below has any relative extrema. Find the values(s)of any relative extrema.
G(x)=(x-3)^2/x-4.
How do I solve this?
To find the x-values of all points where the function G(x) has any relative extrema, you need to find the critical points of the function.
Critical points occur where the derivative of the function is equal to zero or is undefined. So, let's start by finding the derivative of G(x):
G(x) = (x-3)^2 / (x-4)
To find the derivative, you can use the quotient rule:
dG(x)/dx = [(x-4)(2(x-3)) - (x-3)^2(1)] / (x-4)^2
Simplifying the derivative, we get:
dG(x)/dx = [2(x-4)(x-3) - (x-3)^2] / (x-4)^2
Next, we set the derivative equal to zero and solve for x:
2(x-4)(x-3) - (x-3)^2 = 0
Expanding and simplifying, we get:
2(x^2 - 7x + 12) - (x^2 - 6x + 9) = 0
2x^2 - 14x + 24 - x^2 + 6x - 9 = 0
x^2 - 8x + 15 = 0
Factoring the quadratic equation, we get:
(x - 5)(x - 3) = 0
Setting each factor equal to zero, we find the critical points:
x - 5 = 0 or x - 3 = 0
Solving these equations, we get:
x = 5 or x = 3
So, the critical points of G(x) are x = 5 and x = 3.
To find the value(s) of any relative extrema, you can substitute these critical points back into the original function G(x):
For the critical point x = 5:
G(5) = ((5-3)^2) / (5-4)
G(5) = 2^2/1
G(5) = 4
For the critical point x = 3:
G(3) = ((3-3)^2) / (3-4)
G(3) = 0/(-1)
G(3) = 0
Therefore, the function G(x) has a relative maximum at x = 5 with a value of 4, and a relative minimum at x = 3 with a value of 0.
To find the x-values where the function has relative extrema, you need to find the critical points of the function. The critical points occur where the derivative of the function is equal to zero or undefined.
Let's find the derivative of the function G(x):
G'(x) = [(x-4)(2(x-3)) - (x-3)^2(1)] / (x-4)^2
Simplifying G'(x):
G'(x) = [2(x-4)(x-3) - (x-3)^2] / (x-4)^2
G'(x) = [2(x^2 - 7x + 12) - x^2 + 6x - 9] / (x-4)^2
G'(x) = [2x^2 - 14x + 24 - x^2 + 6x - 9] / (x-4)^2
G'(x) = [x^2 - 8x + 15] / (x-4)^2
Now, let's find the critical points by setting the numerator equal to zero:
x^2 - 8x + 15 = 0
(x - 3)(x - 5) = 0
This equation has two solutions: x = 3 and x = 5.
So, the critical points of the function are x = 3 and x = 5.
To find the values of the relative extrema at these critical points, substitute these values into the original function G(x):
For x = 3:
G(3) = (3-3)^2 / (3-4) = 0/(-1) = 0
For x = 5:
G(5) = (5-3)^2 / (5-4) = 2/1 = 2
So, the function has a relative minimum at x = 3 (value = 0) and a relative maximum at x = 5 (value = 2).