# How do you use the first and second derivatives to sketch h(x)=x³-3x+1?

Dec 12, 2016

First derivative will give us critical points:

Points that represent potential local maximum/minimum values or aymptotes

Plugging in values on either side of these critical points will show how the curve is acting (ie. positive/negative slopes)

Second derivative will give us inflection points:

Points in which the curve changes concavity

...As well as tell us if the curve is concave up/down.

#### Explanation:

So, lets first take the derivative of $f \left(x\right)$

$f ' \left(x\right) = 3 {x}^{2} - 3$

$f ' \left(x\right) = 0 \implies 3 {x}^{2} - 3 = 0$

$3 {x}^{2} = 3 \implies x = \pm 1$ Critical values

Check the intervals around the critical values:

For $\left(- \infty , - 1\right)$, the function is positive $\implies$increasing slope

For $\left(- 1 , 1\right)$, the function is negative $\implies$decreasing slope

For $\left(1 , \infty\right)$, the function is positive $\implies$increasing slope

This should make sense since we started with a cubic function.

These changes in slope indicate local extrema.

Local maximum at $x = - 1$
Local minimum at $x = 1$

Now, lets take the second derivative:

$f ' ' \left(x\right) = 6 x$

$f ' ' \left(x\right) = 0 \implies 6 x = 0$

$x = 0$

Check the intervals around this value.

For $\left(- \infty , 0\right)$, the function is negative $\implies$ concave down

For $\left(0 , \infty\right)$, the function is positive $\implies$ concave up

Thus, there is an inflection point at x=0.

Now that we have all that information, we should gather up some basic information to help us plot.

Let's find some intercepts by plugging in 0 for x into the original function f(x).

$f \left(0\right) = {0}^{3} - 3 \left(0\right) + 1$

y-intercept $= 1$

Plug in our extrema values and plot those.

$f \left(- 1\right) = {\left(- 1\right)}^{3} - 3 \left(- 1\right) + 1$

$\left(- 1 , 3\right)$

$f \left(1\right) = {\left(1\right)}^{3} - 3 \left(1\right) + 1$

$\left(1 , - 1\right)$

Now, we know how the graph acts based on those intervals:

Increasing from (-oo, -1)&(1, oo)

Decreasing from $\left(- 1 , 1\right)$

Now concavity; since we know that the function is differentiable along the $\mathbb{R}$

Concave down from $\left(- \infty , 0\right)$
Concave up from $\left(0 , \infty\right)$

graph{x^3-3x+1 [-5.304, 5.796, -2, 3.546]}