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 of a function within a finite domain, you can do:
find f'(x) and equate to zero to find the relative maxima and minima.
In this case, f'(x)=0 at x=1 and x=3.
By checking with f"(x) or otherwise, you will find that x=1 is a minimum, and x=3 is a maximum.
Evaluate f(3)=2ln(12)-3 = 1.97 approx.
Now evaluate function f(x) at limits of domain, namely f(-3)=2ln(12)+3=7.97 approx., and f(5)=2ln(28)-5=1.66 appro.
Thus the absolute maximum is at x=-3.
To find the absolute maximum value of the function f(x) = 2ln(x^2+3) - x in the given domain [-3,5], you'll need to follow these steps:
Step 1: Calculate the critical points of the function.
To find the critical points, you need to determine where the derivative of f(x) is equal to zero or undefined. Start by taking the derivative of f(x) and set it equal to zero:
f'(x) = 0
To find the derivative of f(x), you'll need to use the chain rule and the derivative of ln(x):
f'(x) = 2 * [(1 / (x^2 + 3)) * 2x] - 1
= (4x / (x^2 + 3)) - 1
Now, set f'(x) equal to zero and solve for x:
(4x / (x^2 + 3)) - 1 = 0
Multiply both sides by (x^2 + 3) to eliminate the fraction:
4x - (x^2 + 3) = 0
Simplify the equation:
x^2 - 4x + 3 = 0
Factorize the quadratic equation:
(x - 1)(x - 3) = 0
So the critical points are x = 1 and x = 3.
Step 2: Calculate the endpoints of the domain.
Since the domain is [-3, 5], the endpoints of the interval are x = -3 and x = 5.
Step 3: Evaluate the function at the critical points and endpoints.
Evaluate f(x) at x = -3, x = 1, x = 3, and x = 5. You'll get four values: f(-3), f(1), f(3), and f(5).
Step 4: Compare the function values.
Compare the function values obtained in Step 3 to determine the absolute maximum value. The largest of these values will be the absolute maximum.
Let's calculate the function values to determine the maximum:
f(-3) = 2ln((-3)^2 + 3) - (-3)
= 2ln(9 + 3) + 3
= 2ln(12) + 3 ≈ 5.38
f(1) = 2ln(1^2 + 3) - 1
= 2ln(1 + 3) - 1
= 2ln(4) - 1 ≈ 2.39
f(3) = 2ln(3^2 + 3) - 3
= 2ln(9 + 3) - 3
= 2ln(12) - 3 ≈ 3.77
f(5) = 2ln(5^2 + 3) - 5
= 2ln(25 + 3) - 5
= 2ln(28) - 5 ≈ 4.47
Comparing these values, we see that the greatest value is approximately 5.38, which occurs at x = -3. Therefore, the absolute maximum value of the function is approximately 5.38.