anonymous
  • anonymous
How do you graph this equation manually? 3x^3 - 9x + 5 ≥ 0 Please give me very detailed hints. DX!! I tried to tackle this problem in different ways but obviously, they all failed. ;-;
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
.Sam.
  • .Sam.
First you gotta find the x-intercepts
.Sam.
  • .Sam.
We can find that by equating 3x^3 - 9x + 5 ≥ 0
.Sam.
  • .Sam.
Do you know any approximation methods, like Newton's method?

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.Sam.
  • .Sam.
Because I don't think that will factor out nicely...
anonymous
  • anonymous
;_; I agree, factoring will have a difficult answer. And nope, what is Newton's method? D: Hopefully, you can explain it in laymen terms or like your talking to a 5 year old? :'D
whpalmer4
  • whpalmer4
Some points of interest for sketching the graph can also be found by taking the first derivative and setting that equal to 0. Solve for \(x\).
.Sam.
  • .Sam.
Yeah, if the zeros isn't necessary then just find the turning points will do
anonymous
  • anonymous
What is a derivative? I haven't gone through calculas or my teacher hasn't introduced it in Algebra 2 or Pre-cal
whpalmer4
  • whpalmer4
If nothing else, you can evaluate at the turning points and see where the bumps end up, so to speak :-)
anonymous
  • anonymous
It's getting late and I have to sleep but ill be looking forward to see your seemingly detailed reply in the morning whpalmer4 :D Thanks for helping an inept like me ^^!!!
whpalmer4
  • whpalmer4
Ah, okay, if you haven't had them, then you can't really be expected to use that to help. Here's a quick, trivial explanation: the first derivative is the slope of the curve at a point. At a point where the curve changes direction (like at the bottom of the parabola \(y = x^2\)) the first derivative will = 0. For functions where you just have polynomial terms in powers of \(x\), you can find the derivative easily enough: the derivative with respect to \(x\) of \(ax^n\) is just \(nax^{n-1}\), and the derivative of a bunch of terms is just the sum of the derivatives of the individual terms. Example: \(y=x^2\) Derivative is \(2x\), which = 0 at x = 0. That means that x = 0 is going to be a point worth plotting precisely on your graph, and you should make sure it is on your graph. Part of the game in sketching graphs is knowing where to look more closely! More complicated example: \(y = x^3-3x + 5\) (similar, but not identical to yours) Here the derivative is \(3x^2-3\). Set that equal to 0, \(3x^2-3=0\), \(x^2-1=0\), \(x=\pm1\). I've attached a plot of both the original function and its first derivative (derivative is in purple). Have a look and notice how interesting features line up.
anonymous
  • anonymous
Thanks! :'D I know how to find the derivative now but for some reason I can see how it can help me find the original graph. D: Sorry, could you try explaining that again?
anonymous
  • anonymous
can't see*
anonymous
  • anonymous
:'(
whpalmer4
  • whpalmer4
Here are a few ways in which knowing the derivative of the function you are graphing can help you: 1) finding local minima and maxima If you have a function f(x) to be graphed, and you know its first derivative, commonly notated f'(x), you can set f'(x) = 0 and find the values of x that make that true. These values are points where the instantaneous slope of f(x) = 0, which means either the top of an inverted bowl (for example: \(y = -x^2, x = 0\)), the base of a bowl (\(y = x^2, x=0\)), or an inflection point (\(y = x^3, x = 0\)). I've attached a graph of those three functions so you can see what they look like.
1 Attachment
whpalmer4
  • whpalmer4
2) estimating the slope If you have that first derivative f'(x), you can plug in an arbitrary x value and see how steep your curve is at that point. If you have a handful of points scattered about on your graph, but don't really know how the curve is shaped between them, picking some points in between and computing the first derivative there will give you an idea of the shape of the curve. If the first derivative is +, the line is sloping upward as you go to the right; the larger the slope, the steeper the ascent. 3) verifying the shape If you take the second derivative (repeat the process you used to find the first derivative, except this time do it to the first derivative), you can find if the curve is concave up or down at a given point. If f''(x) is +, then the function is concave up (looks like a U), and if it is -, the function is concave down (inverted U).
anonymous
  • anonymous
1) I don't really understand this part. So, if we have the derivative, does this mean we're able to find an x value that makes the f(x) = 0, which is basically the minimum or maximum values of the original function or derivative? For the example you gave earlier that is similar to mine, I don't really see the connection. D: 2) So, it's basically table of values but using the derivative? 3.) Ah, I think, in laymen terms, this means that I can find it if the function is up or down? If so, I think I get part 3. xD So doing all this won't get me the actual graph but give me an idea of what the actual graph looks like right? If so, for some reason, my brain won't take advantage of the clues and help me piece together the example you gave me that is similar to mine. ToT
Jhannybean
  • Jhannybean
really good explanation @whpalmer4
whpalmer4
  • whpalmer4
For 1): having the 1st derivative and setting it equal to 0 allows us to find the exact points on the curve of the original function where it stops going up, or stops going down. For example, a typical physics problem has you throw a baseball into the air, or fire a cannon, and with all the simplifications made by the textbook authors (no air resistance, etc.), the path turns out to be a parabola, and typically you're asked to find out where the darn thing comes down, and what the maximum height it achieved was. Typically the equation looks something like \(h(t) = -\frac{1]{2}gt^2 + vt + c\) where \(g\) is the acceleration due to gravity, \(v\) is the launch velocity, and \(c\) is a constant representing the initial height (these textbook authors like launching rockets off the tops of buildings and cliffs, never in the middle of a open field like anyone who wants to get their rocket back does). To find the time of flight isn't hard. You set \(h(t) = 0\) and solve for \(t\). Because it is a quadratic, you get two solutions, only one of which may make sense considering the problem (the rocket isn't going to land before it is launched, but you often get a solution involving a negative value of \(t\)). Finding the maximum height is harder, even if you're willing to plot a lot of points on your graph — how do you know a point just to the left or right of the point your eye tells you is the top isn't just a bit higher? Here's where that first derivative comes in handy. If we take the first derivative of \(h(t) = -\frac{1}{2}gt^2 + vt + c\), we get \(h'(t) = -gt + v\). If we set that equal to 0 and solve for \(t\), we get the exact value of \(t\) where the rocket is at its maximum height, which we can then find by plugging that value of \(t\) into our formula for \(h(t)\). Similarly, we can mark that spot on our graph without having to make a table of dozens and dozens of points. We just take \[h'(t) = -gt + v = 0\]\[gt = v\]\[t= v/g\]and plug whatever that works out to be into our original equation to find the maximum height.
whpalmer4
  • whpalmer4
2) it's not really a table, but you could use it to make a table. Think of a rollercoaster: as you start pulling out of the loading area, you go up a steep ramp (f'(x) > 0), until you reach the top (f'(x) = 0), then you plunge down the hill (f'(x) < 0), and at the bottom (f'(x) = 0) you swoop up again (f'(x) > 0), etc.
whpalmer4
  • whpalmer4
3) Yes, you can quickly get an idea of what the function is doing at some arbitrary spot. Generally, the higher the maximum exponent in your function, the more bumps there will be in the curve. For example, \(y = x-x^3/6+x^5/120-x^7/5040+x^9/362880\) — can you quickly tell me what that curve is doing at \(x = -2\)? That's going to be hard to do by making a table!
whpalmer4
  • whpalmer4
That function is a Taylor series approximation to a sine curve, and the graph looks like this:
1 Attachment
whpalmer4
  • whpalmer4
If we take the first derivative, we get \[y'(x) = 1-\frac{1}{2}x^2+\frac{1}{24}x^4-\frac{1}{720}x^6+\frac{1}{40320}x^8\] Second derivative would be \[y''(x) = -x+\frac{1}{6}x^3-\frac{1}{120}x^5+\frac{1}{5040}x^7\] Still a bit challenging to evaluate without a calculator, to be sure! But let's march across the x axis from x = -5 to x = 5 by steps of 1 and see what those values are. \[y(x) = -0.09, 0.7, -0.1, -0.9, -0.8, 0, 0.8, 0.9, 0.1, -0.7, 0.09\]\[y'(x) = 3., -0.4, -1., -0.4, 0.5, 1., 0.5, -0.4, -1., -0.4, 3.\]\[y''(x) =-5., -1., 0.09, 0.9, 0.8, 0, -0.8, -0.9, -0.09, 1., 5.\] Looking at \(y''(x)\) we see that our curve is initially concave down, very much so, but near x = -3, it changes and becomes concave up until x = 0 where it changes again becoming concave down until about x = 3 where it changes again becoming concave up. Looking at \(y'(x)\) we see that the curve has a local top around x = -4 and x = 2, and local bottoms somewhere between x=-2 and x=-1, and near x = 4. Now go back and look at the curve and see how we predicted all of it!
whpalmer4
  • whpalmer4
Looking back at my original graph, the purple parabola (y'(x) = 3x^2-3) crosses the x-axis twice, at x=-1, and x=1. If you extend a vertical line up from those points, you'll see that the blue curve (y(x) = x^3-3x+5) changes direction at those two points. Also, at the vertex of the parabola, the slope of the parabola (which is also the second derivative) is 0, so that represents an inflection point in our original function, where it changes from being concave down to concave up.
anonymous
  • anonymous
Thanks a lot. :') I think derivatives in Calculus will be more bearable because of you. ;D But, for this problem; f(x) = x^5 - 3x^3 - x^2 - 4x - 1, I got the first derivative as f(x) = 5x^4 - 9x^2 - 2x and I don't know how to solve for x at this point. Do I use the 2nd derivative instead?

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