# How do you use the first and second derivatives to sketch f(x)= x^4 - 2x^2 +3?

Sep 20, 2016

The curve is concave downwards at $\left(0 , 3\right)$
The curve is concave upwards at $\left(1 , 2\right)$
The curve is concave upwards at $\left(- 1 , 2\right)$

#### Explanation:

Given -

$y = {x}^{4} - 2 {x}^{2} + 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} - 4 x$

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

To sketch the graph, we have to find for what values of $x$ the slope becomes zero. It means at those points the curve turns.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 4 {x}^{3} - 4 x = 0$

$4 {x}^{3} - 4 x = 0$
$4 x \left({x}^{2} - 1\right) = 0$
$4 x = 0$
$x = 0$
${x}^{2} - 1 = 0$
$x = \pm \sqrt{1}$

$x = 1$
$x = - 1$

The curve turns when x=0; x=1; x=-1

At these points we have to decide whether the curve is concave upwards or concave downwards. For this we need the second derivatives -

At $x = 0$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 12 \left({0}^{2}\right) - 4 = - 4 < 0$

Since the second derivative is less than zero, the curve is concave downwards at $x = 0$

The value of the function is -

$y = {0}^{4} - 2 \cdot {0}^{2} + 3 = 3$

The curve is concave downwards at $\left(0 , 3\right)$

At $x = 1$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 12 \left({1}^{2}\right) - 4 = 8 > 0$

Since the second derivative is greater than zero, the curve is concave upwards at $x = 0$

The value of the function is -

$y = {1}^{4} - 2 \cdot {1}^{2} + 3 = 2$

The curve is concave upwards at $\left(1 , 2\right)$

At $x = - 1$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 12 \left(- {1}^{2}\right) - 4 = 8 > 0$

Since the second derivative is greater than zero, the curve is concave upwards at $x = - 1$

The value of the function is -

$y = {\left(- 1\right)}^{4} - 2 \cdot {\left(- 1\right)}^{2} + 3 = 2$

The curve is concave upwards at $\left(- 1 , 2\right)$