f(x)=2ln(x^2+3)-x Domain [-3,5]
how do i find the absolute maximum value of this?
To find the absolute maximum value of the function f(x) = 2ln(x^2+3) - x within the given domain [-3, 5], you can follow these steps:
1. Find the critical points: The critical points occur where the derivative of the function is either zero or undefined. In this case, the function f(x) is differentiable everywhere within the given domain, so we need to find where its derivative equals zero.
First, let's differentiate f(x) with respect to x:
f'(x) = (2/(x^2+3)) * (2x) - 1
To find the critical points, we set f'(x) equal to zero:
(2/(x^2+3)) * (2x) - 1 = 0
Simplifying this equation, we get:
4x/(x^2+3) - 1 = 0
Multiplying through by (x^2+3), we get:
4x - (x^2+3) = 0
Rearranging terms, we have:
x^2 - 4x + 3 = 0
Factoring this quadratic equation, we get:
(x - 1)(x - 3) = 0
So, the critical points are x = 1 and x = 3.
2. Evaluate the function at the endpoints of the given domain: In this case, the given domain is [-3, 5]. So, we need to evaluate the function at x = -3 and x = 5.
f(-3) = 2ln((-3)^2+3) - (-3)
= 2ln(9+3) + 3
= 2ln(12) + 3
f(5) = 2ln(5^2+3) - 5
= 2ln(25+3) - 5
= 2ln(28) - 5
3. Compare the function values: Now that we have the critical points and the function values at the endpoints, we compare these values to find the absolute maximum.
Evaluate f(x) at the critical points:
f(1) = 2ln(1^2+3) - 1
= 2ln(4) - 1
f(3) = 2ln(3^2+3) - 3
= 2ln(12) - 3
Compare these values with the function values at the endpoints, f(-3) and f(5). The absolute maximum value of the function will be the larger of these values.
Compare:
f(-3), f(1), f(3), f(5)
Whichever of these is the largest will be the absolute maximum value of the function within the given domain.