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

1 Answer
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(x)#

#f'(x) = 3x^2-3#

#f'(x)=0=>3x^2-3=0#

#3x^2=3=>x=+-1# Critical values

Check the intervals around the critical values:

For #(-oo, -1)#, the function is positive #=>#increasing slope

For #(-1, 1)#, the function is negative #=>#decreasing slope

For #(1, oo)#, the function is positive #=>#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''(x)=6x#

#f''(x)=0=>6x=0#

#x=0#

Check the intervals around this value.

For #(-oo, 0)#, the function is negative #=># concave down

For #(0, oo)#, the function is positive #=># 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(0)=0^3-3(0)+1#

y-intercept #=1#

Plug in our extrema values and plot those.

#f(-1)=(-1)^3-3(-1)+1#

#(-1, 3)#

#f(1)=(1)^3-3(1)+1#

#(1, -1)#

enter image source here

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

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

enter image source here

Decreasing from #(-1, 1)#

enter image source here

Now concavity; since we know that the function is differentiable along the #RR#

Concave down from #(-oo, 0)#
Concave up from #(0, oo)#

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