# sketching curve graph

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• May 4th 2013, 12:17 PM
MarkFL
Re: sketching curve graph
Okay good, you have found:

x-intercepts: (-2,0), (-1,0), (1,0)

y-intercept: (0,-2)

Now, can you answer parts a) - c) above? Can you use both the first and second derivative tests to clearly show what the nature of the extrema are? While one one of these test is sufficient, I think is is good in this case to look at both just for an understanding of both methods.

In the end, you will have two extrema, a point of inflection, and 4 intercepts, for a total of 7 points to plot...you will know where the function is increasing/decreasing, and where the function is concave up and concave down. You should be able to make a good sketch of the original function from all of this information.
• May 4th 2013, 12:35 PM
noork85
Re: sketching curve graph
ok, i plotted all the points how do i graph the critical numbers from the first derivative test. they were roots.
this is what my graph looks like so far

https://www.dropbox.com/s/12du0m48t1...2016.34.54.jpg is it correct?
• May 4th 2013, 12:59 PM
noork85
Re: sketching curve graph
• May 5th 2013, 12:46 AM
MarkFL
Re: sketching curve graph
Quote:

Originally Posted by noork85

I apologize for the delay in getting back to you...severe storms in my area knocked out my power for most of the night.

You have the correct intercepts and point of inflection labeled, but your relative minimum is not the y-intercept, and as far as plotting the extrema, use your calculator to get decimal approximations.

Can you first just state what you have found for parts a) - c)? Being able to state these will help you sketch the graph, and I highly recommend that we get those nailed down first before attempting to sketch the graph.
• May 5th 2013, 08:53 AM
noork85
Re: sketching curve graph
i hope all is well....

ok so heres what i coud answer from a-c. i dont understand which points are my local max/min. i mean, i dont know where im supposed to look for them.

https://www.dropbox.com/s/v3v51fukz5...2012.52.01.jpg
• May 5th 2013, 09:11 AM
noork85
Re: sketching curve graph
• May 5th 2013, 09:50 AM
noork85
Re: sketching curve graph
here's what i did for the final graph...includes all my work. not sure about concavity though
https://www.dropbox.com/s/7p5hq991lc...2013.39.34.jpg
https://www.dropbox.com/s/iibkdwpc0j...2013.39.44.jpg
https://www.dropbox.com/s/u91syuxyk1...2013.39.53.jpg (not the best looking graph)
• May 5th 2013, 12:56 PM
MarkFL
Re: sketching curve graph
I didn't have much time (power was out again), but in looking over all of your work, my only point of contention is the intervals of increasing/decreasing behavior. You included zero, and this is not one of your critical values.

I have to run now, but when I get back, I will show you how I would work this problem, from start to finish. :D
• May 5th 2013, 01:07 PM
noork85
Re: sketching curve graph
ok great, thanks.
• May 5th 2013, 03:43 PM
MarkFL
Re: sketching curve graph
We are given:

$f(x)=x^3+2x^2-x-2$

a) find where $f(x)$ is increasing/decreasing.

To do this, we need to compute the first derivative of $f(x)$ and look at where it is positive/negative by equating it to zero.

$f'(x)=3x^2+4x-1=0$

The discriminant is:

$\Delta=(4)^2-4(3)(-1)=28$

This is not a perfect square, so the roots are irrational, therefore factoring is not a practical option for finding the roots, hence we apply the quadratic formula:

$x=\frac{-4\pm\sqrt{28}}{2(3)}=\frac{-4\pm2\sqrt{7}}{2(3)}=\frac{-2\pm\sqrt{7}}{3}$

I would simply observe that the first derivative is a parabola opening upwards to determine it must be negative in between the roots, and positive on either side, thus:

On the interval:

$\left(-\infty,\frac{-2-\sqrt{7}}{3} \right)$ we find $f(x)$ is increasing.

$\left(\frac{-2-\sqrt{7}}{3},\frac{-2+\sqrt{7}}{3} \right)$ we find $f(x)$ is decreasing.

$\left(\frac{-2+\sqrt{7}}{3},\infty \right)$ we find $f(x)$ is increasing.

b) where it is concave up/concave down.

To discuss concavity, we need to look at the second derivative of $f(x)$, in particular it's sign.

$f''(x)=6x+4=0$

Solving for $x$, we find the critical value is at:

$x=-\frac{2}{3}$

Since the second derivative is an increasing linear function, we know then that:

On the interval:

$\left(-\infty,-\frac{2}{3} \right)$ we find $f(x)$ is concave down.

$\left(-\frac{2}{3},\infty \right)$ we find $f(x)$ is concave up.

Because the second derivative does change its sign across the critical value, we may conclude there is a point of inflection at:

$\left(-\frac{2}{3},f\left(-\frac{2}{3} \right) \right)=\left(-\frac{2}{3},-\frac{20}{27} \right)$

c) local max/min

We know from part a) that we have stationary points at:

i) $\left(\frac{-2-\sqrt{7}}{3},f\left(\frac{-2-\sqrt{7}}{3} \right) \right)=\left(\frac{-2-\sqrt{7}}{3},\frac{14\sqrt{7}-20}{27} \right)$

ii) $\left(\frac{-2+\sqrt{7}}{3},f\left(\frac{-2+\sqrt{7}}{3} \right) \right)=\left(\frac{-2+\sqrt{7}}{3},\frac{-14\sqrt{7}-20}{27} \right)$

First derivative test:

Because $f(x)$ is increasing to the left of i) and decreasing to the right, we may conclude that i) is a local maximum.

Because $f(x)$ is decreasing to the left of ii) and increasing to the right, we may conclude that ii) is a local minimum.

Second derivative test:

Since $\frac{-2-\sqrt{7}}{3}<-\frac{2}{3}<\frac{-2-\sqrt{7}}{3}$

We know then that:

$f''\left(-2-\sqrt{7}}{3} \right)<0$ and so i) is a local maximum.

$f''\left(-2+\sqrt{7}}{3} \right)>0$ and so ii) is a local minimum.

d) sketch the curve of the graph.

We know we have:

a local maximum at:

$\left(\frac{-2-\sqrt{7}}{3},\frac{14\sqrt{7}-20}{27} \right)\approx(-1.5485837703548635,0.6311303094409)$

a local minimum at:

$\left(\frac{-2+\sqrt{7}}{3},\frac{14\sqrt{7}+20}{27} \right)\approx( 0.21525043702153024,-2.1126117909223803)$

a point of inflection at:

$\left(-\frac{2}{3},-\frac{20}{27} \right)\approx( -0.6666666666666666,-0.7407407407407407)$

To find the $x$-intercepts, equate $f(x)=0$ and find the roots:

$f(x)=x^3+2x^2-x-2=x^2(x+2)-(x+2)=$

$(x^2-1)(x+2)=(x+2)(x+1)(x-1)=0$

So, we know the $x$-intercepts are at:

$(-2,0),\,(-1,0),\,(1,0)$

To find the $y$-intercept, we let $x=0$ and find:

$f(0)=-2$

and so we know the $y$-intercept is at:

$(0,-2)$

Putting all of this together, we should obtain a graph that looks like:

Attachment 28257
• May 5th 2013, 04:07 PM
noork85
Re: sketching curve graph
thank you so much. i understand it so much better now. i was afraid i would fail the graphing part of my exam tomorrow, but im fairly confident i can do this. thank u again!!!
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