# How do you sketch the graph y=x^4-27x using the first and second derivatives?

May 25, 2017

#### Explanation:

As $y = {x}^{4} - 27 x$

observe that ${x}^{4} - 27 x = x \left({x}^{3} - 27\right)$ and as ${x}^{4}$ is always positive and rises faster than $27 x$, curve, it is negative only between $0 < x < 3$ and elsewhere it is positive.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} - 27$ and $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 12 {x}^{2}$

Now $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ when $4 {x}^{3} - 27 = 0$

i.e. ${\left(\sqrt[3]{4} x\right)}^{3} - {3}^{3} = 0$

or $\left(\sqrt[3]{4} x - 3\right) \left({\left(\sqrt[3]{4} x\right)}^{2} + 3 \sqrt[3]{4} x + 9\right) = 0$

or $\left(\sqrt[3]{4} x - 3\right) \left(\sqrt[3]{16} {x}^{2} + 3 \sqrt[3]{4} x + 9\right) = 0$

And $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ when $\sqrt[3]{4} x - 3 = 0$ or $x = \frac{3}{\sqrt[3]{4}} = 1.8899$

and as at this point $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0$, we have a minima.

And for root(3)16x^2+3root(3)4x+9)=0, discriminant is $9 \sqrt[3]{16} - 4 \times 9 \sqrt[3]{16} = - - 27 \sqrt[3]{16} < 0$ and no other solution for $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ exists.

graph{x^4-27x [-10, 10, -60.2, 99.8]}