the eqn of curve is x³ - x² - 5x-3
dy/dx=3x² -2x-5
d²y/dx²=6x-2
how to show that gradient of the curve is never less than -16/3
Assume that the gradient of the curve COULD be less than $\displaystyle -\frac{16}{3}$.
Then $\displaystyle \frac{dy}{dx} = 3x^2 - 2x - 5 < -\frac{16}{3}$
$\displaystyle 3x^2 - 2x + \frac{1}{3} < 0$
$\displaystyle 3\left(x^2 - \frac{2}{3}x + \frac{1}{9}\right) < 0$
$\displaystyle 3\left[x^2 - \frac{2}{3}x + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \frac{1}{9}\right] < 0$
$\displaystyle 3\left[\left(x - \frac{1}{3}\right)^2 - \frac{1}{9} + \frac{1}{9}\right] < 0$
$\displaystyle 3\left(x - \frac{1}{3}\right)^2 < 0$
This is clearly a contradiction, as $\displaystyle \left(x - \frac{1}{3}\right)^2 > 0$.
Therefore the gradient can never be less than $\displaystyle -\frac{16}{3}$.