Graphing Functions of Differentiation

[frosty8688]

Homework Statement

Find the intervals of increase or decrease.
$$h(x)=(x+1)^{5}-5x-2$$

Homework Equations

The Attempt at a Solution

I found the derivative to be
\begin{align*}
h'(x) &= 5(x+1)^{4}-5 \\
&= 5[(x+1)^{4}-1] \\
&= (x^{4}+4x^{3}+6x^{2}+4x+1-1) \\
&= (x^{4}+4x^{3}+6x^{2}+4x) \\
&= x(x^{3}+4x^{2}+6x+4) \\
&= x[x^{2}(x+4)+2(3x+2)]
\end{align*} That is what I got so far.

 

[jbunniii]

If your goal is to find out where $h$ is increasing or decreasing, then your first expression for $h'(x)$ is probably easier to work with than the rest of the stuff you wrote. For what values of $x$ is the following inequality true?
$$5(x+1)^4 - 5 >0$$

 

[frosty8688]

When x > 0 and when x < -2.

 

[jbunniii]

When x > 0 and when x < -2.

Please explain how you got this answer.

 

[frosty8688]

When x is greater than 0, the first part of the equation is greater than 5. Let's say you put in 1 for x and 5*2 - 5 = 5 which is greater than 0. Also, when x is less than -2, let's use -3, then -2^4 = 16 and 5*16= 80 and 80-5= 75, so x is greater than 0.

 

[jbunniii]

When x is greater than 0, the first part of the equation is greater than 5. Let's say you put in 1 for x and 5*2 - 5 = 5 which is greater than 0. Also, when x is less than -2, let's use -3, then -2^4 = 16 and 5*16= 80 and 80-5= 75, so x is greater than 0.

It seems like you are just plugging in values for $x$, as opposed to solving the inequality. A more careful line of argument might start as follows:

Observe that
$$5(x+1)^4 - 5 > 0$$
if and only if
$$5(x+1)^4 > 5$$
if and only if
$$(x+1)^4 > 1$$
If we put $y = x+1$ then this is equivalent to
$$y^4 > 1$$
Since the equation $y^4 = 1$ has exactly two real solutions ($y = \pm 1$), it follows that the equation $(x+1)^4 = 1$ has exactly two real solutions ($x = 0, -2$).

Now how can we use this to obtain the desired conclusion?

 

[frosty8688]

The book has 0 and -2 as critical values.

 

[jbunniii]

The book has 0 and -2 as critical values.

Thanks, I edited my previous post to fix the sign error. So what can you say about the sign of $h'(x)$ in each of the following intervals: $(-\infty, -2)$, $(-2, 0)$, $(0, \infty)$?

 

[frosty8688]

It is increasing on (-∞,-2) and (0,∞) and decreasing on (-2,0) because x is greater than 0 on the intervals (-∞,-2) and (0,∞) and less than 0 on (-2,0).

 

[jbunniii]

Looks right to me.

 

[HallsofIvy]

The derivative in this case is continuous. Once you have found the "critical points", where the derivative is 0, the derivative can change from "+" to "-" only where it is 0 so "plugging in values of x" in each interval is a perfectly valid way of solving the inequality.

 

[frosty8688]

The derivative in this case is continuous. Once you have found the "critical points", where the derivative is 0, the derivative can change from "+" to "-" only where it is 0 so "plugging in values of x" in each interval is a perfectly valid way of solving the inequality.

I found that to be easier, because I still couldn't find out from the derivative how -2 was a critical number.

 

[SteamKing]

What's the value of the derivative when x = -2?

 

[frosty8688]

What's the value of the derivative when x = -2?

The answer is 0.