# How do you use the first and second derivatives to sketch y= 3x^4 - 4x^3?

May 23, 2018

#### Explanation:

Given: $f \left(x\right) = y = 3 {x}^{4} - 4 {x}^{3} = {x}^{3} \left(3 x - 4\right)$

Find the first derivative and set $y ' = 0$ to find the critical points:

$\text{ } y ' = 12 {x}^{3} - 12 {x}^{2} = 0$

$\text{ } y ' = 12 {x}^{2} \left(x - 1\right) = 0$

$\text{ }$critical values: $\text{ } x = 0 , x = 1$

$\text{ "f(0) = 0; " } f \left(1\right) = 3 {\left(1\right)}^{4} - 4 {\left(1\right)}^{3} = - 1$

$\text{ }$critical points: $\left(0 , 0\right) , \left(1 , - 1\right)$

Find the second derivative and evaluate it at the critical values to find relative minimums or relative maximums:

$y ' ' = 36 {x}^{2} - 24 x$

$y ' ' \left(0\right) = 0 \text{ } \implies$need to use the first derivative test to find out if this point is a relative min., max. or inflection.

$y ' ' \left(1\right) = 36 {\left(1\right)}^{2} - 24 \left(1\right) = 12 > 0 \implies$relative min. at $\left(1 , - 1\right)$

Find inflection points by setting $y ' ' = 0$:

$y ' ' = 36 {x}^{2} - 24 x = 0$

$y ' ' = 12 x \left(3 x - 2\right) = 0 \implies$ inflections at $x = 0 , x = \frac{2}{3}$

$f \left(\frac{2}{3}\right) = 3 {\left(\frac{2}{3}\right)}^{4} - 4 {\left(\frac{2}{3}\right)}^{3} = 3 \cdot \frac{16}{81} - 4 \cdot \frac{8}{27} = - \frac{16}{27}$

inflection points $\left(0 , 0\right) , \left(\frac{2}{3} , - \frac{16}{27}\right)$

To graph:

Fourth order functions have end conditions both positive. Put a point at $\left(0 , 0\right)$ and a point at the relative minimum $\left(1 , - 1\right)$

Start at the upper left and create a horizontal inflection at $\left(0.0\right)$, then turn downward to the relative minimum, then curve upward.

graph{3x^4-4x^3 [-5, 5, -5, 5]}