Question

Consider the function \( f(x) = -x^2 + 10x + 100 \). The minimum value of the function in the interval \( [5, 10] \) is \(\underline{\hspace{2cm}}\).

Hint:

To find the minimum of a quadratic function, check the critical points and evaluate at the endpoints of the interval.

Answer 1

Correct Answer: 100

To find the minimum value of \( f(x) = -x^2 + 10x + 100 \) in the interval \( [5, 10] \), we first compute the derivative: \[ f'(x) = -2x + 10 \] Setting \( f'(x) = 0 \) to find the critical points: \[ -2x + 10 = 0 $\Rightarrow$ x = 5 \] Now, we evaluate \( f(x) \) at the endpoints \( x = 5 \) and \( x = 10 \): \[ f(5) = -(5)^2 + 10(5) + 100 = -25 + 50 + 100 = 125 \] \[ f(10) = -(10)^2 + 10(10) + 100 = -100 + 100 + 100 = 100 \] Thus, the minimum value is \( 100 \).