$10$ The equation of a curve is $y=4 x^{3}+4 p x^{2}+16 x-9$ . Find the range of values of $p$ such that $y$ is an increasing function.

Full solution

10

The equation of a curve is

y=4 x^{3}+4 p x^{2}+16 x-9

. Find the range of values of

p

such that

y

is an increasing function.

Find Derivative of y: To find when $y$ is increasing, we need to find the derivative of $y$ with respect to $x$ , which gives us the slope of the tangent to the curve at any point. $\newline$ $y' = \frac{d}{dx} (4x^3 + 4px^2 + 16x - 9)$ $\newline$ $y' = 12x^2 + 8px + 16$
Check Derivative for Increase: For $y$ to be increasing, its derivative $y'$ must be greater than or equal to $0$ for all $x$ . So, we need to find the values of $p$ such that $12x^2 + 8px + 16 \geq 0$ for all $x$ .
Calculate Discriminant: The expression $12x^2 + 8px + 16$ is a quadratic in $x$ . For this quadratic to be non-negative for all $x$ , its discriminant must be less than or equal to $0$ . $\newline$ Discriminant, $D = (8p)^2 - 4(12)(16)$ $\newline$ $D = 64p^2 - 768$
Set Discriminant Inequality: Set the discriminant less than or equal to $0$ to find the range of $p$ . $64p^2 - 768 \leq 0$
Simplify Inequality: Divide by $64$ to simplify the inequality. $p^2 - 12 \leq 0$
Analyze Parabola Interval: This inequality represents a parabola that opens upwards, and we want the values of $p$ within the interval where the parabola is below the x-axis. $(p - \sqrt{12})(p + \sqrt{12}) \leq 0$
Identify Increasing y Values: The roots of the equation $p^2 - 12 = 0$ are $p = \sqrt{12}$ and $p = -\sqrt{12}$ . So, the values of $p$ for which $y$ is increasing are in the interval $[-\sqrt{12}, \sqrt{12}]$ .