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Prove the following.
Prove that y = 1/2(-8x² + 4x + 5) is an increasing function in (-∞, 1/4].
Prove that $y = \frac{1}{2}(-8x^2 + 4x + 5)$ is an increasing function in $\left(-\infty, \frac{1}{4}\right]$.
Given: $y = \frac{1}{2}(-8x^2 + 4x + 5)$
To show: $y$ is an increasing function in $\left(-\infty, \frac{1}{4}\right]$.
Answer
Answer
$y = \frac{1}{2}(-8x^2 + 4x + 5)$
$\frac{dy}{dx} = \frac{1}{2}(-16x + 4) = -8x + 2 = 2(1 - 4x)$
For $x \le \frac{1}{4}$: $4x \le 1$, so $1 - 4x \ge 0$ and $\frac{dy}{dx} \ge 0$ (zero only at $x = \frac{1}{4}$).
Hence $y$ is an increasing function in $\left(-\infty, \frac{1}{4}\right]$.
- 1. $y = \frac{1}{2}(-8x^2 + 4x + 5)$ (given)
- 2. $\frac{dy}{dx} = \frac{1}{2}(-16x + 4) = 2(1 - 4x)$ (differentiating term by term)
- 3. For $x \le \frac{1}{4}$, $1 - 4x \ge 0$, so $\frac{dy}{dx} \ge 0$, with equality only at the end point $x = \frac{1}{4}$
- 4. A function whose derivative is $\ge 0$ on an interval (zero only at isolated points) is increasing there
- 5. $y$ is an increasing function in $\left(-\infty, \frac{1}{4}\right]$.