Question

Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).

Hint:

For a function to be increasing, its derivative must be positive. Find the value of \( a \) that satisfies this condition on the given interval.

Answer 1

Step 1: For the function \( f(x) = x^2 + ax + 1 \) to be increasing, its derivative must be positive on the interval \( [1, 2] \). Step 2: Compute the derivative: \[ f'(x) = 2x + a \] Step 3: For the function to be increasing, we need \( f'(x)>0 \). Thus: \[ 2x + a>0 \] For \( x = 1 \), we get: \[ 2(1) + a>0 \quad \Rightarrow \quad a>-2 \] Step 4: Since \( a>-2 \) satisfies the condition for increasing on the interval \( [1, 2] \), the least value of \( a \) is \( a = -2 \).