Thread: how to prove that gradient of curve is never less than something

1. how to prove that gradient of curve is never less than something

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

2. Originally Posted by helloying
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 $-\frac{16}{3}$.

Then $\frac{dy}{dx} = 3x^2 - 2x - 5 < -\frac{16}{3}$

$3x^2 - 2x + \frac{1}{3} < 0$

$3\left(x^2 - \frac{2}{3}x + \frac{1}{9}\right) < 0$

$3\left[x^2 - \frac{2}{3}x + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \frac{1}{9}\right] < 0$

$3\left[\left(x - \frac{1}{3}\right)^2 - \frac{1}{9} + \frac{1}{9}\right] < 0$

$3\left(x - \frac{1}{3}\right)^2 < 0$

This is clearly a contradiction, as $\left(x - \frac{1}{3}\right)^2 > 0$.

Therefore the gradient can never be less than $-\frac{16}{3}$.