200 + 8x^3 + x^4 and find the interval of increase. To do this one, the first thing to do is to take the derivative.
Now, a derivative is nothing but the slope of a curved line. That means that, wherever the line is flat, the derivative is equal to 0. This also means that anything that ONLY affects on the position of the line on the y-axis, because slope is a proportion, doesn't change the value of the slope. Therefore, I can eliminate 200, giving me:
8x^3 +x^4
Now, on the face of it, you take a derivative in the same way that you would a slope. Now, the way that you would calculate a slope for a straight line would to simply make this little computation: y(#2) - y(#1) divided by x(#2) - x(#2). This would give us a slope equal to the difference in y divided by the difference in x.
Now, in calculus, we think of y as a "function of x." What this means is that when we say "y = x^2" then you can ALWAYS substitute x^2 in for y. This means that you could theoretically type x^2 - x^2 divided by x-x. The problem with this theory is that you can't divide anything by zero because it gives you an undefined number.
Therefore, we're going to use an algebraic trick to get around this problem. We're going to add another variable to x wherever it appears in the equation, giving us instead (x+h)^2 - x^2 divided by x+h -x. The denominator simplifies to h.
Simplifying further, we break down the binomial x+h squared into x^2 + 2xh + h^2 to give us a numerator of x^2 + 2xh + h^2 - x^2
x^2 eliminates, giving us 2xh + h^2 divided by h. Factor out the h, and you get 2x + h.
Now, by putting in this extra variable, I have simplified the expression considerably by getting the fraction out of the picture. However, now I have a problem: I didn't really want to have that h there!
That's okay. Since I only really intended h as an algebraic place-holder, I can just say "Oh, well it was always equal to 0 anyway," thereby leaving me with simply 2x. Therefore, I have taken the limit of my original function as the difference between x (#1) and x(#2) approached the value of 0.
Back to my equation: 8x^3 + x^4
Okay, gonna throw in a few rules.
1) if you add together two slopes, you get the slope of the two functions you got them from. This means that, because, when you add together 2x and 3x, you get the sum of their slopes times x, you can always assume that, when you add two functions together, you add up the slopes.
Therefore, let's find the slope of 8x^3 first. Now, we know that a slope is 0 when a function is flat. We also know that a slope is negative when a function is "going down from left to right." Well, we know that any function cubed increases exponentially on one side and decreases exponentially on the other, so we can assume that the function is flat in the middle and steep on the sides.
We can also assume that it is increasing from right to left on both sides, right?
Therefore, the slope will be positive on both sides...right?
Therefore, what we are looking for is an equation that is close to 0 in the middle and positive on both sides.
Let's say it's x^2, then.
Now, here is another thing that we know: the higher the exponent, the more steeply it rises. Well, doesn't this mean that a larger exponent means a larger slope?
Well, let's say we did some algebra and found out that the slope of x^3 is equal to 3x^2. Well, hell, that makes sense, doesn't it? That works with the theory that a larger exponent begets a larger slope.
However, it's all well and good to say that the slope of our function will look something like the slope of x^3, but that doesn't tell us what to do with the constant, 8, in the slope of the expression 8x^3. I know how to solve that little problem, though.
Let's act out an example, here, using a linear equation:
We know the slope of 5x is 5. Well, let me use this as an example. The slope of x is 1, and we know that the constant 5 gives us a straight line therefore a slope of 0.
Solve this little equation: (0 * x) + (5 * 1) That is "the slope of 5 times x" plus "5 times the slope of x."
(0 * x) + (5 * 1) = 5. Now let's try it with the equation we're trying to work with.
The slope of 8 is 0.
The slope of x^3 is 3x^2
(0 * x^3) + (8 * 3x^2) = 24x^2
Therefore, the derivative of 8x^2 is the coefficient 8 times the exponent times x^(exponent minus 1)
So let's just take it on faith that the same rules would work with x^4
Derivative of (that is the "slope of the function") 8x^3 + x^4 is 24x^2 + 4x^3
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Now, to find the interval of increase, I just work out where my function is greater than 0.
24x^2 + 4x^3 = 0
x^2 * (24 + 4x) = 0
0 * any number is equal to zero, so the functions can be divided into x^2 = 0, 24 + 4x = 0
x = -6, 0
x = -5 means 24*(-5)^2 + 4*(-5)^3=100 positive
x = -7 gives -196
x = 1 gives 28
This is as expected because, with this type of equation where you have the highest exponent odd and the other exponent one less, it sort of "kisses" the x-axis at 0. Graph as many of these as you want to, but it's always going to be positive on both sides of 0. The other x value is less than 0, and the coefficient of the other function in the equation greater than 0. Below the line it goes.
Intervals of increase are (-6,0) and (0,ifni)
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I want the maximum, so I have to find where the slope is 0 and negative on both sides. That isn't the case anywhere, though, because I'm looking at a negative, a positive, and then a positive. That means that the shape of my line goes down, stops, rises, stops, and then rises again. There really isn't a maximum val there.
On the other hand, I do have a minimum because I have a place on the line where the slope is 0 and positive on both sides, which is at x = -6. Well, this is easy because, in my original equation, 200 + 8x^3 + x^4, plugging -6 in for x gives me -232.
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Okay, concavity, concavity. That means I want to know where the slope stops increasing and starts decreasing or the reverse of that. I can determine that with a second derivative.
Derivative of 24x^2 + 4x^3 = 48x + 12x^2
Stops increasing or decreasing at 48x + 12x^2 = 0
x(48 + 12x)
x = 0
x = -4
Curve's shaped generally like a quadratic, crosses through the x-axis twice, positive on both sides, irregularity at -4 that I could probably take a third derivative and say stuff about, I'm pretty sure it's concave downward between -4 and 0.
What this of course means is that the slope stops increasing and starts decreasing at -4, which means that it starts to go in the direction of flattening out from being negative. Then at 0 it switches again and starts increasing.
Essentially, this means that our base equation has a sort of wiggle or "s-shape" between -6 and 0, where it's sort of WANTING to go down again but doesn't quite manage to move any in that direction before it crosses over point 0 and starts getting steeper again.
++++++++++++++++++++++
200 + 8x^3 + x^4
Okay, to get inflection points, I plug in my 0s for the second derivative (into my original equation), and I of course get 200 for point 0 giving me (0, 200) and for -4, let's see, 4 cubed is equal to 8 squared, and we get 8 cube after multiplying with the coefficient, getting 2 cubed cubed or 2^9 and negative because the coefficient is negative and the exponent odd, and then 4^4 is 2 squared^4 or 2^(2*4) getting 2^8. 0 minus the difference is -2^8 which is -256. That subtracted from 200 is -56.
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