To find the global minimum of the function \(f(x) = 8 \log_e x - x^2 + 3\) over the interval \([1, e]\), we need to follow a systematic approach involving calculus. This method typically includes finding critical points and evaluating the function at these points, as well as at the endpoints of the given interval.

Function and Interval Definition

The given function for which we need to find the global minimum is:

\[f(x) = 8 \log_e x - x^2 + 3\]

The specified closed interval for this optimization problem is:

\[[1, e]\]

Here, \(\log_e x\) represents the natural logarithm of \(x\), and we are provided with the approximate value of \(e^2 \approx 7.39\) for calculations involving the endpoint \(e\).

Derivative Calculation of f(x)

The first step in finding the global minimum is to calculate the first derivative of the function \(f(x)\) with respect to \(x\). This derivative, \(f'(x)\), helps us identify points where the function's slope is zero, which are potential locations for local minima or maxima.

Let's differentiate \(f(x)\):

\[f(x) = 8 \log_e x - x^2 + 3\]

Using the standard rules of differentiation from calculus:

• The derivative of \(c \cdot \log_e x\) is \(c \cdot \frac{1}{x}\).
• The derivative of \(x^n\) is \(n x^{n-1}\).
• The derivative of a constant (like \(3\)) is \(0\).

Applying these rules, we obtain the derivative \(f'(x)\) as:

\[f'(x) = \frac{d}{dx}(8 \log_e x) - \frac{d}{dx}(x^2) + \frac{d}{dx}(3)\]

\[f'(x) = 8 \cdot \frac{1}{x} - 2x + 0\]

\[f'(x) = \frac{8}{x} - 2x\]

Critical Points Identification

To find the critical points of the function within the specified interval, we set the first derivative \(f'(x)\) equal to zero and solve for \(x\). These points are crucial because an extremum (minimum or maximum) can occur here.

Set \(f'(x) = 0\):

\[\frac{8}{x} - 2x = 0\]

To solve for \(x\), we can rearrange the equation:

\[\frac{8}{x} = 2x\]

Multiply both sides by \(x\) (note that \(x\) must be greater than 0 for \(\log_e x\) to be defined, so \(x \neq 0\)):

\[8 = 2x^2\]

Divide both sides by 2:

\[x^2 = \frac{8}{2}\]

\[x^2 = 4\]

Taking the square root of both sides gives us two possible values for \(x\):

\[x = \pm \sqrt{4}\]

\[x = \pm 2\]

Now, we need to check which of these critical points lie within our given interval \([1, e]\). We know that \(e \approx 2.718\).

• For \(x = 2\): Since \(1 \le 2 \le e\) (meaning \(1 \le 2 \le 2.718\)), \(x=2\) is indeed within the interval.
• For \(x = -2\): This value is not within the interval \([1, e]\) because the interval starts from \(1\).

Therefore, the only relevant critical point for finding the global minimum in this problem is \(x=2\).

Evaluating Function at Endpoints and Critical Points

To determine the global minimum of the function over the closed interval \([1, e]\), we must evaluate the original function \(f(x)\) at the endpoints of the interval (\(x=1\) and \(x=e\)) and at the identified critical point (\(x=2\)). The smallest value among these will be the global minimum.

Let's calculate \(f(x)\) for each of these points:

Determining the Global Minimum

After evaluating the function at all relevant points (the endpoints and the critical point within the interval), we compare the calculated \(f(x)\) values to identify the global minimum.

• The value of the function at the left endpoint, \(f(1) = 2\).
• The value of the function at the critical point, \(f(2) \approx 4.544\).
• The value of the function at the right endpoint, \(f(e) = 3.61\).

Comparing these three values (\(2\), \(4.544\), and \(3.61\)), the smallest value is \(2\). This minimum value occurs when \(x = 1\).

Therefore, the function \(f(x) = 8 \log_e x - x^2 + 3\) attains its global minimum over the interval \([1, e]\) at \(x = 1\).