Exercise 6.2 · Q4: Where $2x^2-3x$ increases and decreases

Basics you need

Solution

1. Differentiate

   By the sum and power rules,

   $$f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(3x) = 4x - 3.$$

   $f$ is a polynomial, so it is continuous and differentiable everywhere: Theorem 1 can be used on any closed interval.

   The direction of $f$ at each point is the sign of its slope $f'(x)$⁠. So the whole question reduces to: where is $4x - 3$ positive, and where negative?
   $\frac{d}{dx}(x^n) = nx^{n-1}$ and the derivative of a sum is the sum of the derivatives (Section 5.3, Table 5.3).

2. Find where $f'(x) = 0$

   $$4x - 3 = 0 \iff x = \frac34.$$

   The point $x = \frac34$ divides the real line into two disjoint intervals, $\left(-\infty, \frac34\right)$ and $\left(\frac34, \infty\right)$⁠.

   A continuous $f'$ cannot go from negative to positive without passing through 0 (the intermediate value property of continuous functions, which this book uses without proof). So on each interval that contains no root of $f'$⁠, the sign of $f'$ is the same throughout. Here we do not even need that fact: Step 3 reads the sign of $4x - 3$ directly for every $x$⁠.
   Solving a linear equation: add 3, divide by 4. Basics: factorising and the discriminant

3. Sign of $f'$ on each interval

   • If $x < \frac34$⁠, then $4x < 3$⁠, so $f'(x) = 4x - 3 < 0$⁠. (Test point: $f'(0) = -3$⁠.)
   • If $x > \frac34$⁠, then $4x > 3$⁠, so $f'(x) = 4x - 3 > 0$⁠. (Test point: $f'(1) = 1$⁠.)
   We argued the sign for every $x$ in each interval (multiply $x < \frac34$ by 4), not just at the test point. The test points are a quick check that the chart has the right signs.
   Multiplying an inequality by a positive number keeps its direction: $x < \frac34 \Rightarrow 4x < 3$⁠. Basics: sign of a product

4. Conclude with Theorem 1

   On $\left(-\infty, \frac34\right)$⁠: for any $x_1 < x_2$ in it, $f' < 0$ on $(x_1, x_2)$⁠, so Theorem 1 (b) and its proof give $f(x_1) > f(x_2)$⁠. Hence $f$ is (strictly) decreasing on $\left(-\infty, \frac34\right)$.

   On $\left(\frac34, \infty\right)$⁠: in the same way $f' > 0$ gives $f(x_1) < f(x_2)$⁠, so $f$ is (strictly) increasing on $\left(\frac34, \infty\right)$.

   This matches the shape: $2x^2 - 3x$ is an upward parabola with vertex at $x = -\frac{b}{2a} = \frac{3}{4}$⁠, so it must fall before the vertex and rise after it.

5. What about the point $x = \frac34$ itself?

   $f$ is continuous at $\frac34$⁠. Theorem 1 asks only for continuity at the ends of $[a, b]$⁠, so it applies on $\left[\frac34, b\right]$ for every $b > \frac34$⁠, and on $\left[a, \frac34\right]$ for every $a < \frac34$⁠. So we may also say $f$ is increasing on $\left[\frac34, \infty\right)$ and decreasing on $\left(-\infty, \frac34\right]$⁠. Both forms are correct. The book's own Example 10 writes open intervals, and so does the answer key.

   No larger interval works: an interval containing some $a < \frac34$ and some $b > \frac34$ contains $\left[a, \frac34\right]$⁠, on which $f$ strictly falls, so $f$ is not increasing on it, and similarly not decreasing.

   Including the turning point is a question about the definition, not about the derivative: at $\frac34$ the derivative is 0, but continuity lets the point join either neighbouring interval. Example 12 in the book does exactly this, closing its intervals at the turning point "by Theorem 1".
   Square bracket = endpoint included; round bracket = excluded. Basics: interval notation

Every fact used, and where it comes from