if the derivative of f is f'(x)=ln(x^2+1)-2, at which of the following values of x does f have an absolute minimum value?
2.53
0
4.44 or
1.72
To find the value of x at which the function f has an absolute minimum, we need to find the critical points of the function. A critical point occurs when the derivative of the function is equal to zero or undefined.
Given that the derivative of f is f'(x) = ln(x^2 + 1) - 2, we need to find when f'(x) = 0.
Setting f'(x) equal to zero:
ln(x^2 + 1) - 2 = 0
To solve this equation, first, add 2 to both sides:
ln(x^2 + 1) = 2
Next, convert the equation from logarithmic to exponential form:
e^(ln(x^2 + 1)) = e^2
This simplifies to:
x^2 + 1 = e^2
Subtracting 1 from both sides gives:
x^2 = e^2 - 1
Taking the square root of both sides yields:
x = ± sqrt(e^2 - 1)
Now, we need to determine the values of sqrt(e^2 - 1). Evaluating this expression gives approximately:
sqrt(e^2 - 1) ≈ 2.7289
So, the critical points are x = -2.7289 and x = 2.7289.
Next, we need to determine whether these critical points correspond to an absolute minimum or maximum. Since f'(x) = ln(x^2 + 1) - 2, we can analyze the concavity of the function.
Calculating the second derivative f''(x):
f''(x) = (d/dx) (ln(x^2 + 1) - 2)
= (2x) / (x^2 + 1)
To determine if a critical point corresponds to a minimum or maximum, we check whether f''(x) is positive or negative.
For x = -2.7289:
f''(-2.7289) = (2 * -2.7289) / ((-2.7289)^2 + 1) < 0
For x = 2.7289:
f''(2.7289) = (2 * 2.7289) / ((2.7289)^2 + 1) > 0
Based on the sign of f''(x) at each critical point, we can conclude that x = 2.7289 corresponds to an absolute minimum value for f.
Therefore, the answer is 2.7289. None of the other values (2.53, 0, 4.44, or 1.72) satisfy the condition for an absolute minimum value.