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

**helloying** · September 18th 2009, 07:33 PM

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

**Prove It** · September 18th 2009, 07:53 PM

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 $-\frac{16}{3}$ .

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