# Homework Help: Intervals of increase and decrease

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1. Mar 20, 2015

### cptstubing

1. The problem statement, all variables and given/known data
Find the intervals of increase/decrease. Use first derivative test to find the local maxima and minima. Sketch a rough graph.

2. Relevant equations
a) f(x) = 2x2+12x-1
b) f(x) = 1/2x4-2x2
c) f(x) = 3x-4 sqrt 2
d) f(x) = x2-1/x2+1

3. The attempt at a solution
A)
it becomes f '(x) = 4x+12
solve for zero
4x+12=0
4x=-12
x=-3

plug x=-3 into the equation to get y value, which is 0
so...
(-3, 0) is the first and only critical point.
After testing it, i've determined that
f(x) decreases when xE (-infinity, -3)
f(x) increases when xE (-3, infinity)
This is correct I believe, but I wouldn't be surprised if it's not.

B) I end up with critical points (0,0) and (sqrt2, -2)

My attempt is:
f(x) = 1/2x4-2x2
f '(x) = 2x3-4x
solve for zero
2x3-4x
------is this next step right?
2x(x2-2) = 0
2x=0
x=0
and....
x2-2=0
x2=2

so after plugging 0 and sqrt2 into original equation give me (0,0) and (sqrt2, -2)
Intervals in the end area negative, another negative, and a positive value.

C) the derivative of this would be 3, which leaves no x value to determine.
D) quotient rule to find derivative, which I think is 4x/(x2+1)2

I think I need a math tutor, this course is killing me.

Last edited: Mar 20, 2015
2. Mar 20, 2015

### Ray Vickson

Substituting the value of x into f(x) will not tell you anything at all about whether x is a max, a min, or a saddle point. You need to examine behavior on intervals.

First, here is a simple "trick" that helps analyze a polynomial $f(x) = a x^n + b x^{n-1} + \cdots + r$, with leading coefficient $a > 0$.
(i) If $n$ is even (2,4,6,...) then $f$ is decreasing for large negative $x$ and increasing for large positive $x$. (Basically, the graph of $y = f(x)$ behaves somewhat like the parabola $y = x^2$, and if you look at the latter graph you will see what I mean.)
(ii) If $n$ is odd (1,3,5,...) then $f(x)$ is increasing for both positive and negative $x$; just look at the graph of $y = x^3$ to see what I mean.

The reason why this simple trick works is that for very large $|x|$ the leading term $a x^n$ swamps the other, lower-order terms, at least in terms of general behavior.

If $a < 0$, just reverse the directions (i.e, change "increasing" to "decreasing", etc.) Basically, this follows just by looking at $-f(x)$, whose leading coefficient would again be > 0.

Here is how I would do the first question.
(a) $f(x) = 2 x^2 + 12 x - 1$, so the derivative is $f'(x) = 4 x + 12$. The only stationary point is $x = -3$, as you have found. The trick above tells you that $f$ decreases in $(-\infty,-3)$ and increases in $(-3,\infty)$.

You can do the other three yourself. Of course, the function in (d) is not a polynomial, but if you think about it you can still say what its behavior must be for large positive and negative $x$, as well as for small positive and negative $x$.

Last edited: Mar 20, 2015
3. Mar 20, 2015

### SammyS

Staff Emeritus
First of all, this and at least one of your previous threads belong in the Calculus homework Forum. This is pre-calculus.

As for part (a) :

You found the critical x value just fine, but you plugged that back into the derivative function, not into the function, f(x), itself, so you didn't actually find the minimum.​

As for some other issues:

You need to be more systematic in approaching these problems. You seem to be going in three directions at once.

Clearly your algebra skills need improvement. That's all too common with many Calculus students.

I suspect that what you mean for part (d) is f(x) = (x2-1)/(x2+1) .

What you wrote literally means: f(x) = x2 - (1/x2) +1 . That's a much different function. So use enough parentheses to communicate what you intend to communicate .

$\displaystyle f(x)=\frac{x^2-1}{x^2+1}\$ versus $\displaystyle \ f(x)=x^2-\frac{1}{x^2}+1$

4. Mar 20, 2015

### cptstubing

Do you mean (see bold above) 'increases in $(-3,\infty)$?

5. Mar 20, 2015

### cptstubing

This course is labelled Mathematics 12 - Precalculus. There is a calculus course as well but I don't need it.
I need my math mark upgraded and it's proving to be a task, not that I didn't anticipate a challenge.

6. Mar 20, 2015

### Staff: Mentor

As soon as you start using derivatives, you are using calculus to do so, regardless of how your school labels the course.

7. Mar 20, 2015

### RUber

For (b) don't forget the negative value that satisfies $x^2 =2$.
For (d), since the denominator is always positive, you should be able to learn a lot by characterizing the numerator.

8. Mar 20, 2015

### Ray Vickson

Yes, of course. Typo fixed.

Anyway, what is your answer to people who asked you about what you meant in question (d)? Exactly what is the function $f(x)$? When I offered a hint, I was assuming the function was $f(x) = x^2 - \frac{1}{x^2} + 1$. If that is not true, the hint would be misleading!