Q.Find the value of the following: For what values of $a$ the function $f$ given by $f(x) = x^2 + ax + 1$ is increasing on $[1, 2]$?

Concept: Derivative Sign Analysis

A function is increasing on an interval when its derivative is non-negative throughout that interval.

Step 1 – Find the derivative

$f'(x) = 2x + a$.

Step 2 – Condition for increasing on $[1,2]$

We need $f'(x) \ge 0$ for all $x \in [1,2]$.

Since $f'(x)$ is linear and increasing in $x$, the smallest value on $[1,2]$ occurs at $x = 1$.

Step 3 – Apply the condition …

The key idea is that a function is increasing on an interval if its derivative is non-negative throughout that interval. For a smooth function like a quadratic, the derivative tells us the slope at every point. If the slope never dips below zero on $[1,2]$, the function is rising (or at least not falling) as we move right.

Here, $f(x) = x^2 + ax + 1$ is a parabola opening upwards (coefficient of $x^2$ is $1 > 0$). Such a parabola decreases until its vertex, then increases after. So the function will be increasing on $[1,2]$ exactly when the entire interval $[1,2]$ lies to the right of the vertex. That is, the vertex’s $x$-coordinate must be $\leq 1$.

Let’s work through it step by step.

1. Find the derivative.

   $f'(x) = 2x + a$.

   This is a linear function — its sign changes at the point where $f'(x) = 0$, i.e., at $x = -a/2$. That point is the vertex of the parabola.

2. Condition for increasing on $[1,2]$.

   For $f$ to be increasing on $[1,2]$, we need $f'(x) \geq 0$ for every $x$ in $[1,2]$. Since $f'(x)$ is linear, its minimum on a closed interval occurs at one of the endpoints. So it’s enough to check the endpoints: if $f'(1) \geq 0$ and $f'(2) \geq 0$, then $f'(x) \geq 0$ everywhere in between.

   Tip

   For a linear function, the sign on an interval is determined entirely by the signs at the endpoints. No need to check every point.

3. Apply the endpoint conditions.

   • At $x = 1$: $f'(1) = 2(1) + a = 2 + a \geq 0 \implies a \geq -2$.
   • At $x = 2$: $f'(2) = 2(2) + a = 4 + a \geq 0 \implies a \geq -4$.

   The stricter condition is $a \geq -2$ (since $-2 > -4$). So $a \geq -2$ guarantees both endpoints are non-negative.

4. Check the vertex interpretation. …

Method: Finding Parameter Values That Make a Function Increasing on a Closed Interval

Use this method whenever a question gives a family of functions depending on an unknown parameter (like $a$) and asks for which values of the parameter the function is increasing (or decreasing) on a stated closed interval $[p,q]$.

Steps

Step 1: Differentiate with respect to $x$, treating the parameter as a constant.

This gives $f'(x)$ as an expression that still contains the parameter.

Step 2: Write down the condition for "increasing on $[p,q]$": $f'(x) \ge 0$ for every $x$ in $[p,q]$.

The non-strict inequality is used because "increasing" on a closed interval only requires the derivative to be non-negative, not strictly positive at isolated points.

Step 3: Exploit the shape of $f'(x)$ to reduce "for every $x$" to a check at the interval's endpoints.

If $f'(x)$ is linear in $x$ (as it is whenever $f$ itself is a quadratic), its minimum value on a closed interval occurs at one of the two endpoints — so it suffices to require $f'(p)\ge 0$ and $f'(q)\ge 0$, then keep whichever resulting condition on the parameter is stricter. …

Common Mistakes

Mistake 1: Using the strict inequality $f'(x) > 0$ instead of $f'(x) \ge 0$.

Why it's wrong: "increasing" in this context only requires the derivative to be non-negative, with equality allowed at isolated points such as a single endpoint — a strict inequality unnecessarily excludes the boundary value of the parameter. Correct approach: use $\ge 0$ unless the question explicitly demands strict increase with no flat point anywhere.

Mistake 2: Checking only one endpoint of the interval instead of both.

Why it's wrong: for a linear derivative, the minimum over a closed interval could occur at either endpoint — skipping one risks missing the binding (stricter) condition. Correct approach: evaluate $f'(x)$ at both endpoints and keep the more restrictive resulting inequality. …