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So, we got a function for example, $y=\dfrac{2x^2}{x+2}$.

I have to find extreme values and intervals of increasing (decreasing).

So... we start off by finding $f'(x)$. Right.. once we find it, we have to equal it to $0$. So, $f'(x)=0$. Once we find values for $x$ to be $0$, we set our points to be $M_1 (x_1,y_1)$ and $M_2 (x_2,y_2)$

Then we continue to find $y_1$, $y_2$ by replacing the values of $x_1$,$x_2$ we found earlier on the main function.

And we get the two points...

Ok, I hope I'm doing good so far.

Next to do is to find the $f''(x)$. Once we find it, we can see which one is $<0$ and which one is $>0$, for example if, $f''(x_1)>0 - \min ($concave$)$, and if $f''(x)<0 - \max ($convex$)$.

Am I doing it right?

mardat
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user2041143
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2 Answers2

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You're doing pretty good. Let me propose you an alternative, which does not require to compute $f''$.

Since you computed $$ f'(x)=\frac{2x^2+8x}{(x+2)^2}=\frac{2x(x+4)}{(x+2)^2}\quad\forall x\neq -2 $$ you don't need much more work to get all your answers.

Observe that $f'$ is positive on $(-\infty,-4)\cup(0,+\infty)$, so $f$ is increasing there.

Then $f'$ is negative on $(-4,-2)\cup(-2,0)$, so $f$ is decreasing there.

Note that $\{x=-2\}$ is a vertical asymptote.

Now write down these variations from $-\infty$ to $+\infty$: draw an increasing arrow where $f$ is increasing, and a decreasing one where it is decreasing. You see at once that $f(-4)=-16$ is a local maximum, and that $f(0)=0$ is a local minimum.

Given the behaviour near the asymptote, $\lim_{-2^-}f=-\infty$ and $\lim_{-2^+}f=+\infty$. So the above extrema are local, and not global.

Finally, take your favorite graphing calculator and check your answer. enter image description here

Julien
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Best way to work out this kind of problem is actually doing it step by step, as you did in your post, but more actions involved. It's called mathematical analysis.

1) Find domain of your function. Obviously, all points except $x = -2$ $$ D[f] = \left ( -\infty, -2 \right ) \cup \left (-2, +\infty \right) = \mathbb R \setminus\{-2\} $$

2) Find range of your function. It's also quite obvious that function can take any values. $$ E[f] = (-\infty, +\infty) = \mathbb R $$

3) Find positive-negative range. To do this, consider the function itself. For example $$ f(x) = \frac {2x^2}{x+2} $$ Enumerator is always positive or $0$. Denominator is positive if $x > -2$ and negative if $x < -2$. To illustrate it typically people draw a diagram as follows

posneg

Not that $x = -2$ is excluded from the $x$ line since it doesn't belong to the domain of function.

4) Find asymptotes if any

a) Horizontal or inclined asymptote. To find it, assume that there's one of the form $y = kx+ b$. In this case $$ \lim_{x \rightarrow +\infty} \frac {f(x)}x = k \\ $$ and $$ \lim_{x \rightarrow +\infty} (f(x)-kx) = b $$ should hold. So check it $$ \lim_{x\rightarrow +\infty} \frac{\frac {2x^2}{x-2}}x = \lim_{x\rightarrow +\infty} \frac {2x}{x+2} = 2 \\ \lim_{x\rightarrow +\infty}\left ( \frac {2x^2}{x+2}-2x \right) = \lim_{x\rightarrow +\infty} \frac {-4x}{x+2} = -4 $$ So $y = 2x-4$ is an inclined asymptote of $f(x)$.

Note: Sometimes functions have two different asymptotes when $x\rightarrow +\infty$ and $x\rightarrow -\infty$, so they need to be found separately by calculating the limit above with $x \rightarrow -\infty$ as well. If yo actually do it, you'll find that for this particular $f$ both asymptotes coincide.

b) To find vertical asymptotes one needs to define points $a$ such that $$ \lim_{x\rightarrow a \pm 0}f(x) = \pm \infty $$ Looking at $f$ you can say that $x = -2$ is the only point when limit above holds. $$ \lim_{x\rightarrow a +0} \frac {x^2}{x+2} = +\infty \\ \lim_{x\rightarrow a -0} \frac {x^2}{x+2} = -\infty $$ So $x = -2$ is a vertical asymptote. On the left from it the function is infinitely negative, from the right – infinitely positive.

5) Find extrema and intervals of monotonicity. This part can be done by observing sign of derivative $$ f'(x) = \left (\frac {2x^2}{x+2} \right)' = \frac {2x(x+4)}{(x+2)^2} $$ Diagram for this expressions will be

negpos2

Looking at this diagram you can say that when $x < -4$ function is increasing, when $x \in (-4, -2)\cup (-2, 0)$ function is decreasing, when $x > 0$ it's increasing again, which means at $x = -4$ function has its local maximum and at $x = 0$ its local minimum.

So now you know pretty much everything about functions behaviour on different ranges, so you can sketch it.

plot

Kaster
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