Question

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

• A. $2$
• B. $1$
• C. $e$
• D. $\dfrac{1+e}{2}$

Hint:

Always check boundary points when optimizing a function on a closed interval, even if a stationary point exists inside.

Answer 1

The Correct Option is B

To find the minimum of \[ f(x)=8\ln x - x^2 + 3, \] we compute the derivative: \[ f'(x)=\frac{8}{x} - 2x. \] Set $f'(x)=0$: \[ \frac{8}{x} = 2x \quad \Rightarrow \quad 8 = 2x^2 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = 2. \] Step 1: Check feasibility.
The interval is $[1, e] \approx [1, 2.718]$. Since $x = 2$ lies inside the interval, we evaluate the function at: \[ x = 1,\quad x = 2,\quad x = e. \] Step 2: Evaluate values.
\[ f(1) = 8(0) - 1 + 3 = 2, \] \[ f(2) = 8\ln 2 - 4 + 3 \approx 5.545 - 1 \approx 4.545, \] \[ f(e) = 8 - e^2 + 3 \approx 11 - 7.389 \approx 3.61. \] Minimum value occurs at $x = 1$.
Final Answer: $1$