# Thread: Estimating how large/small expressions are - calculus

1. ## Estimating how large/small expressions are - calculus

I managed the first of this type of question, but am struggling with the next as it involves potentially negative numerator.
Estimate how large and how small the following expression can be. Carefully consider whether each term is positive or negative.

1-3x/x+2 for 0<x<1

2. I assume the expression is $f(x) = \frac{1-3x}{x+2}$. The absolute extrema occur at critical numbers $c$ such that $f'(c) = 0$ or $f'(c)$ is undefined.

$f'(x) = -\frac{7}{(x+2)^2}$

As you can see, there are no critical numbers $c$ such that $f'(c) = 0$. There is a critical number $c$ such that $f'(c)$ is undefined, namely $c = -2$. However, $c = -2$ is not contained in the open interval $(0,1)$. Therefore, there are no critical numbers and, consequently, no extrema.

Note as x approaches 0 from the right, f(x) approaches $\frac{1}{2}$. And as x approaches 1 from the left f(x) approaches - $\frac{2}{3}$. However, I wouldn't consider these the absolute maximum and minimum values of f(x) on the open interval 0 < x < 1 because f(x) does not actually take on those values.

3. I don't think that's the idea. I'll go through how I did question 1 then can you see if you can help?

1) Find how large (2x+2)/ (3x+1) for 0<x<1 can be:

Since x<1; 2x<2, 2x+2<4

Since 0<x; 0<3x 1<3x+1

So 1(2x+2) < 4(3X+1)

And therefore (2x+2)/(3x+1) < 4

But for the question I posted, apparently it can't be done in quite this way as it takes in possible negative values?

Argh damn this is for tomorrow and I have no idea still !

4. So you aren't actually finding the absolute extrema, just the upper and lower bounds. In that case, $-\frac{2}{3} < \frac{1 - 3x}{x + 2} < 1$ if $0 < x < 1$.

I noticed you didn't use calculus to solve that last problem. Are you allowed to use calculus?

For demonstrative purposes, here is a solution to that last problem using calculus.

Let $f(x) = \frac{2x + 2}{3x + 1}$. Then, $f'(x) = -\frac{4}{(3x + 1)^2}$.

Because $f'(x) < 0$ on the interval $0 < x < 1$, $f(x)$ is decreasing on the interval $0 < x < 1$.

In other words, $f(x)$ is becoming smaller and smaller as $x$ approaches $1$ (from the left). The opposite holds true - $f(x)$ is becoming larger and larger as $x$ approaches $0$ (from the right).

Because $\lim_{x \to 0^+}f(x) = 2$ and $\lim_{x \to 1^-}f(x) = 1$, $1 < \frac{2x + 2}{3x + 1} < 2$.

A graph of $f(x)$ confirms my solution.

5. I don't think that's what they are asking. This is only very basic first year stuff as I've had two weeks of my first year at uni.

I have dne more steps than I showed though, using axioms to get from step to step...

6. Here is a graph of $\frac{2x + 2}{3x + 1}$ from x = 0 to x = 1: plot &#40;2x &#43; 2&#41;&#47;&#40;3x &#43; 1&#41; from x&#61;0 to 1 - Wolfram|Alpha. As you can see, the graph ranges from 1 to 2, exclusive. In other words, $1 < \frac{2x + 2}{3x + 1} < 2$.

Your reasoning is correct although your conclusion is inaccurate. $\frac{2x + 2}{3x + 1}$ is indeed less than 4 on the interval $0 < x < 1$. It's simply more accurate to write $1 < \frac{2x + 2}{3x + 1} < 2$. I'm not sure which answer your teacher prefers or if my answer can even be found using your method. I'll give it some more thought and post again soon.