estimate the absolute maximum and minimum values of f if any, on the stated interval, and then use calculus methods to find the exact values:
f(x)=10ln(x^2+1)-8x; [0,3]
To find the absolute maximum and minimum values of f on the interval [0,3], we will use the following steps:
1. Find the critical points of f(x) within the interval [0,3]. These are the points where the derivative of f(x) is either zero or does not exist.
2. Evaluate f(x) at each critical point and at the endpoints of the interval to find the maximum and minimum values.
Step 1: Find the critical points of f(x)
To find the critical points, we need to find where the derivative of f(x) equals zero or does not exist. Let's start by finding the derivative:
f'(x) = d/dx [10ln(x^2+1) - 8x]
= 10 * (1/(x^2+1)) * 2x - 8
= 20x / (x^2+1) - 8
Next, we set the derivative equal to zero and solve for x:
20x / (x^2+1) - 8 = 0
20x - 8(x^2+1) = 0
20x - 8x^2 - 8 = 0
8x^2 - 20x + 8 = 0
2x^2 - 5x + 2 = 0
Now we can factor the quadratic equation:
(2x - 1)(x - 2) = 0
Setting each factor equal to zero, we get:
2x - 1 = 0 --> x = 1/2
x - 2 = 0 --> x = 2
So, the critical points within the interval [0,3] are x = 1/2 and x = 2.
Step 2: Evaluate f(x) at critical points and endpoints
Now, we need to evaluate f(x) at the critical points x = 1/2 and x = 2, as well as at the endpoints x = 0 and x = 3.
f(0) = 10ln((0)^2+1) - 8(0) = 10ln(1) - 0 = 10ln(1) = 0
f(1/2) = 10ln((1/2)^2+1) - 8(1/2) = 10ln(1/4+1) - 4 = 10ln(5/4) - 4
f(2) = 10ln((2)^2+1) - 8(2) = 10ln(4+1) - 16 = 10ln(5) - 16
f(3) = 10ln((3)^2+1) - 8(3) = 10ln(9+1) - 24 = 10ln(10) - 24
Now, we will evaluate f(x) at each point and compare to find the maximum and minimum values.
f(0) = 0
f(1/2) ≈ -7.404
f(2) ≈ -5.194
f(3) ≈ -3.629
From the evaluations, we can observe that f(3) ≈ -3.629 is the smallest value, and f(0) = 0 is the largest.
In conclusion, the absolute minimum value of f on the interval [0,3] is approximately -3.629, which occurs at x = 3.
The absolute maximum value of f on the interval [0,3] is 0, which occurs at x = 0.