find the absolute extrema if they exist as well as where they occur
f(x)= x-1/x^2+3x+5
To find the absolute extrema of a function, we need to find the highest and lowest points on the graph of the function.
To find the absolute extrema of the function f(x) = (x - 1)/(x^2 + 3x + 5), we need to follow these steps:
Step 1: Find the critical points by taking the derivative of the function and solving for when the derivative equals zero or is undefined.
Step 2: Evaluate the function at the critical points and the endpoints of the defined interval to find the highest and lowest values.
Step 1: Find the critical points:
To find the derivative of f(x), we need to use the quotient rule. The quotient rule states that if we have (u/v)', it is equal to (v * u' - u * v') / v^2.
Let's differentiate f(x) = (x - 1) / (x^2 + 3x + 5):
f'(x) = ((x^2 + 3x + 5)(1) - (x - 1)(2x + 3)) / (x^2 + 3x + 5)^2
Next, we can simplify the expression:
f'(x) = (x^2 + 3x + 5 - (2x^2 + x - 3)) / (x^2 + 3x + 5)^2
f'(x) = (-x^2 + 2x + 8) / (x^2 + 3x + 5)^2
Now we need to find the critical points by solving (-x^2 + 2x + 8) / (x^2 + 3x + 5)^2 = 0:
-x^2 + 2x + 8 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In our case, a = -1, b = 2, and c = 8:
x = (-2 ± sqrt(2^2 - 4(-1)(8))) / 2(-1)
x = (-2 ± sqrt(4 + 32)) / -2
x = (-2 ± sqrt(36)) / -2
x = (-2 ± 6) / -2
The two possible solutions for x are x = 4 and x = -2.
Step 2: Evaluate the function at the critical points and endpoints:
Now that we have our critical points, we need to evaluate our function f(x) at these points and the endpoints of the defined interval. However, since no interval is specified, we'll find the absolute extrema for the entire domain.
f(4) = (4 - 1) / (4^2 + 3(4) + 5) = 3 / 37
f(-2) = (-2 - 1) / ((-2)^2 + 3(-2) + 5) = -3 / 15
In addition to the critical points, we need to check the values of the function at the endpoints of the domain. Since there are no specific limits mentioned, we will assume the domain to be all real numbers.
Now we need to determine the behavior of the function as x approaches positive infinity and negative infinity.
When x approaches positive infinity, both the numerator and denominator of f(x) approach infinity, and their ratio approaches 0.
When x approaches negative infinity, both the numerator and denominator of f(x) approach negative infinity, and their ratio approaches 0.
Therefore, there are no absolute extrema for the function f(x) = (x - 1) / (x^2 + 3x + 5) because the function does not have any local maxima or minima.