List the critical numbers of the following function in increasing order. Enter N in any blank that you don't need to use.
f(x) = (6x+6)/(5x^2+5x+5)
x = -1 is a critical number arising from the numerator
(-1,0) is the x-intercept
for 5x^2 + 5x + 5 , nothing to see here.
I guess, we could let x = 0, to get y-intercepts which can be done mentally
y = 6/5
List the critical numbers of the following function in increasing order. Enter N in any blank you don't need
To find the critical numbers of the function f(x) = (6x+6)/(5x^2+5x+5), we need to find the values of x where the derivative of the function is equal to zero or undefined.
Step 1: Find the derivative of the function.
f'(x) = [((6)(5x^2 + 5x + 5) - (5x^2 + 5x + 5)(6))/ (5x^2 + 5x + 5)^2]
= [(30x^2 + 30x + 30 - 30x^2 - 30x - 30)/ (5x^2 + 5x + 5)^2]
= [0 / (5x^2 + 5x + 5)^2]
= 0
Step 2: Set the derivative equal to zero and solve for x.
0 = 0
Since the equation 0 = 0 is always true, there are no critical numbers of the function f(x) = (6x+6)/(5x^2+5x+5).
Therefore, the function f(x) = (6x+6)/(5x^2+5x+5) does not have any critical numbers.
To find the critical numbers of a function, we need to first find the derivative of the function and then solve for the values of x where the derivative is equal to zero or undefined.
Given the function: f(x) = (6x+6)/(5x^2+5x+5)
Step 1: Find the derivative of the function.
To find the derivative of f(x), we can use the quotient rule. The quotient rule states that if we have a function in the form of f(x) = g(x) / h(x), the derivative is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's apply the quotient rule to find the derivative of f(x):
f'(x) = [(6 * (5x^2 + 5x + 5) - (6x + 6) * (10x + 5)) / (5x^2 + 5x + 5)^2]
Now we simplify the expression:
f'(x) = [30x^2 + 30x + 30 - (60x^2 + 60x + 30) / (5x^2 + 5x + 5)^2]
f'(x) = [-30x^2 - 30x - 30 / (5x^2 + 5x + 5)^2]
Step 2: Find the values of x where the derivative is equal to zero or undefined.
To find the critical numbers, we set f'(x) equal to zero and solve for x:
-30x^2 - 30x - 30 = 0
We can simplify the equation by dividing all terms by -30:
x^2 + x + 1 = 0
However, this quadratic equation does not have real solutions. Therefore, there are no critical numbers for this function.
So, in the case of the given function, there are no critical numbers.