Exercise 6.2 · Q14: Values of $a$ making $x^2+ax+1$ increasing on $[1,2]$

Basics you need

Solution

1. Differentiate $f$

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

   For every value of $a$⁠, $f$ is a polynomial, so it is continuous on $[1, 2]$ and differentiable on $(1, 2)$⁠: the conditions of Theorem 1 hold.

   $a$ is a constant, so $\frac{d}{dx}(ax) = a$ and $\frac{d}{dx}(1) = 0$⁠. The whole question is now about the sign of $2x + a$ for $x$ between 1 and 2.
   Power rule and constant multiple rule (Class XI, Limits and Derivatives).

2. When is $2x + a \ge 0$ for every $x \in (1, 2)$⁠?

   If $a \ge -2$⁠: for every $x \in (1, 2)$ we have $2x \gt 2$⁠, so

   $$2x + a \gt 2 + a \ge 0.$$

   So $f'(x) \gt 0$ on all of $(1, 2)$⁠, strictly positive, not just $\ge 0$⁠.

   If $a \lt -2$⁠: then $2 + a \lt 0$⁠, and for $x$ just above 1 the value $2x + a$ is close to $2 + a$⁠, so negative. For example, at $x = 1 + t$ with $t = \min\left(\frac12, \frac{-2-a}{4}\right) \gt 0$⁠, $2x + a = 2 + a + 2t \le \frac{2 + a}{2} \lt 0$⁠.

   So $f'(x) \ge 0$ for every $x \in (1, 2)$ exactly when $a \ge -2$.

   The smallest values of $2x + a$ on $(1, 2)$ are near $x = 1$⁠, where it is close to $2 + a$⁠. So everything hinges on the sign of $2 + a$⁠. Note that at $a = -2$ the value $2 + a = 0$ is only approached at the excluded end point $x = 1$⁠, never reached inside.
   Adding $a$ to both sides of $2x \gt 2$ keeps the inequality: $2x + a \gt 2 + a$⁠. Basics: interval notation

3. For $a \ge -2$⁠, $f$ is increasing on $[1, 2]$

   By Step 1 the hypotheses of Theorem 1 hold, and by Step 2, $f'(x) \gt 0 \ge 0$ for each $x \in (1, 2)$⁠. Theorem 1 (a) gives: for every $a \ge -2$⁠, $f$ is increasing on $[1, 2]$. Because $f'$ is strictly positive on $(1, 2)$⁠, the Mean Value Theorem step in the proof gives $f(x_1) \lt f(x_2)$ for $1 \le x_1 \lt x_2 \le 2$⁠: $f$ is even strictly increasing.

   This is the "sufficient" half of the answer: every $a \ge -2$ works. We still have to show that no other $a$ works, because the question asks for all such values.
   Theorem 1 (a), Section 6.3. Lesson: increasing and decreasing functions

4. For $a \lt -2$⁠, $f$ is not increasing on $[1, 2]$

   Compare any $x$ with $x = 1$⁠. Since $f(1) = 1 + a + 1 = 2 + a$⁠,

   $$\begin{aligned} f(x) - f(1) &= x^2 + ax + 1 - (2 + a)\\ &= (x^2 - 1) + a(x - 1)\\ &= (x - 1)(x + 1 + a). \end{aligned}$$

   Let $a \lt -2$ and take $x_2 = 1 + t$ with $t = \min\left(1, \frac{-2-a}{2}\right)$⁠. Then $t \gt 0$⁠, so $1 \lt x_2 \le 2$⁠, and $x_2 - 1 = t \gt 0$⁠. Also $t \le \frac{-2-a}{2}$⁠, so

   $$\begin{aligned} x_2 + 1 + a &= 2 + a + t\\ &\le 2 + a + \tfrac{-2-a}{2} = \tfrac{2 + a}{2} \lt 0. \end{aligned}$$

   So $f(x_2) - f(1)$ is a positive number times a negative number: $f(x_2) - f(1) \lt 0$⁠. The points $1 \lt x_2$ of $[1, 2]$ have $f(1) \gt f(x_2)$⁠. By Definition 1, $f$ is not increasing on $[1, 2]$ when $a \lt -2$.

   A concrete case: $a = -3$⁠, $x_2 = \frac32$ gives $f(1) = -1$ and $f\left(\frac32\right) = \frac94 - \frac92 + 1 = -\frac54$⁠, and $-1 \gt -\frac54$⁠.

   Theorem 1 only gives sufficient conditions, so "$f'$ is negative somewhere" does not by itself prove "not increasing". A pair of points with $f(x_1) \gt f(x_2)$ does, straight from Definition 1. The factorised form shows the drop: just to the right of 1 the factor $x + 1 + a$ is close to $2 + a \lt 0$⁠.
   $x^2 - 1 = (x - 1)(x + 1)$⁠; a product of a positive and a negative number is negative. Basics: sign of a product

5. The boundary value $a = -2$ directly

   With $a = -2$⁠, $f(x) = x^2 - 2x + 1 = (x - 1)^2$⁠. For $1 \le x_1 \lt x_2 \le 2$ we have $0 \le x_1 - 1 \lt x_2 - 1$⁠, and squaring non-negative numbers keeps the order, so

   $$f(x_1) = (x_1 - 1)^2 \lt (x_2 - 1)^2 = f(x_2).$$

   So for $a = -2$⁠, $f$ is strictly increasing on $[1, 2]$⁠, confirming Step 3 without any theorem.

   This step matters because the answer key leaves $a = -2$ out. Here $f'(1) = 0$⁠, but $1$ is an end point, and a zero derivative at an end point does not stop a function from increasing on the interval.
   If $0 \le p \lt q$ then $p^2 \lt q^2$⁠. Basics: sign of a product

6. Answer, and the answer key

   Steps 3 and 4 together: $f$ is increasing on $[1, 2]$ exactly when $a \ge -2$. The same set works if "increasing" is read as "strictly increasing", since Step 3 gave strict increase for every $a \ge -2$⁠.

   The answer key prints $a \gt -2$⁠. Every value in the key's answer is correct, but it misses $a = -2$⁠, which Step 5 shows does work. The key's answer is what you get by demanding $f'(x) \gt 0$ on the closed interval $[1, 2]$ (that is, also $f'(1) = 2 + a \gt 0$⁠); Theorem 1 asks for the sign of $f'$ only on the open interval $(1, 2)$⁠.

   Whenever an answer is an inequality, test the boundary value separately. Here the boundary value $a = -2$ is the case where the parabola's lowest point lands exactly on the end point $x = 1$⁠.

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