BAM!
Mr. Tarrou!
In this video, I would like to reemphasize why you care where the tangent lines are horizontal.
And you are going to find out where the slope of your function...here we have a sort of weird looking function,
f(x), I've just sketched off a line.
Why do we care about where its slope is equal to zero? Why do we care about where the tangent lines are equal
to zero? Well, I've marked off three points here in this graph.
One here, at value c. One here, at value b.
And one here, at the x-coordinate of a. We care about where the slope is equal to zero...or we
care about where the slope of the tangent line are a equal to zero, because we can then analyze
the derivative to the left and to the right of those values of x where the graph has a slope of zero.
We can see if we found a relative minimum or maximum.
If to the left of this value were the slope is zero, if the first derivative is negative, and to the right the first
derivative is positive, then your graph is falling with a negative slope, then bottoms out, then starts to rise
with a positive slope.
That is a relative minimum.
Or, if to the left of your particular value that you find a slope of zero, or the tangent line is horizontal,
if the derivative is positive to the left and negative to the right, you've found a relative maximum.
And then, just because you find a place where the tangent line has a slope of zero, or the slope
of the curve is equal to zero, you have not automatically found a relative max or min.
Most of the time you will have, but not all the time.
In this case, I have a graph that is falling, so my first derivative will be negative.
Then it just sort of flattens out for a second, and then just happens to start to fall again.
So that means that we can find places in our function where the derivative is equal to zero, yet we have not
found a relative min or max.
So, we are just going to do two examples. I have one with just a normal polynomial, and another one with
trig functions and a little bit of use of the chain rule. We are going to find out where those derivatives are equal
to zero. And that is all we are going to do at that point. But again, once you find out where those derivatives are
equal to zero, you can plug in a number, into the derivative, just a little bit less and a little bit higher,
and look for those sign changes if they do indeed exist. Again, negative on left, positive on right - relative min.
Positive on left and negative on the right of this paricular x-value where the slope is equal to zero - relative max.
No sign change, whether it is going down on both sides or up - no sign change means...not a relative max or min.
Let's see what those examples look like!
WHOO!
Here is our first example. We have f(x) is equal to x-cubed minus 4x-squared,
plus x, plus 6. We are going to find the derivative.
So f'(x) is equal to... we will apply the power rule here and we get 3 times x-squared.
We have a constant in the power, so we are going to bring that down. 2 times negative 4 is negative 8, times x.
Here we have plus x, which is just going to be one. And the constant is going to be zero.
So the derivative is 3x-squared minus 8x plus 1.
This is the derivative. This will tell us the slope of this function at any point where x is defined.
We are going to solve for what? We want the derivative to be zero. That is when the
tangent lines are horizontal. We are going to let f'(x) equal zero.
3x-squared, minus 8x plus 1. And see if we can factor this, because it will take a
little bit less work. We are going to take the first and last term - multiply
those together. Get 3, and look at that product - is there a way to factor 3, the product of the leading coefficient
and the constant - are their factors of 3 that are also going to add up to be negative 8?
The answer to that is no.
So, this trinomial - this quadratic is not factorable. So, that means that we are going to, since it is a quadratic,
solve it using the quadratic formula.
That is, in case you did forget, x is equal to negative b, plus/minus the square root of b-squared,
minus 4ac all over 2a.
And of course we know that we want the equation set equal to zero, we want all of the terms here with the
exponents counting down.
We have "a" is equal to 3, "b" is equal to negative 8, and "c" is equal to 1.
So, doing our substitution, we have x equals the opposite of b, plus/minus the square root of b-squared
minus 4ac, all over 2 times a.
I like to use the parenthesis and lay it out before I even start plugging the numbers in. That way, if I am plugging
in a negative, I will know that there was a negative from the formula, and a negative from the values that I am
plugging in.
So it is the opposite of negative 8, plus/minus the square root of negative 8-squared, minus 4 times a,
which is 3, times c, which is 1.
All over 2 times a.
All we have to do is work this out. I am not going to worry about checking to the left and right to see if
it is a relative max or min. I will leave that up to you, if the problems you are doing requires that.
We are just going to focus on finding the x-values at which the derivative is equal to zero.
We have x=positive 8, plus/minus the square-root of negative 8-squared, which is 64.
Negative 12, from negative 4 times 3. All over 2 times 3, which is equal to 6.
64-12=52. So we have 8 plus/minus the square root of 52, over 6.
Now, I want to leave this in exact form. I want to see if there is a way to reduce the square root of 52.
I am looking for a perfect square in 52, like 4, 9, 16, 25...I am pretty sure from my notes that the biggest
perfect square here is 4. And let me see...yes.
So we are going to rewrite this as 8 plus/minus the square root of 4 times 13. Right?
4 times 10 is 40, and 4 times 3 is 12. 40+12=52.
Over 6. You do not have to, but it is nice to leave the perfect
square first all the time, because that is going to come out. The square root of 4 is equal to 2.
So we have 8 plus/minus 2 times the square root of 13, over 6. And of course, 6 is even, 8 is even.
2 is even. All three of these terms have a factor of 2. They are all even. If there was a one there, and it was
just the 8 and the 6 that were even, nothing would reduce. But, we can factor a 2 out of here.
Factor a 2 out of the 6. Those 2's cancel out... and boom! There is your answer!
4 plus/minus the square root of 13 over 3.
One more example - it will involve trig functions.
WHOO!
For our last example, we have f(x)=2sin(x) + cos(2x). We want to find the derivative, so that means,
that we are going to have f'(x) is equal to...now the derivative of sine is cosine...so we have 2cos(x).
Plus, we have the cosine of 2x. Now, here, I can just write cos(x) and be done finding the derivative of sine.
But here, we have this inside function, u, is equal to 2x. We are going to find the derivative of cosine.
The derivative of cosine... you shouldn't be writing these sign symbols assuming you know what it is going to be.
Let's take that out. The derivative of cosine is negative sine. So, minus sin(2x). But, now I have to take care of
the derivative of the inside function. So this is the derivative of the cosine function, times to derivative of
the inside function. The derivative of 2x is equal to 2.
There is not much here to clean up. We have 2 times cos(x)...I want to take the 2 out of both of these terms.
So, minus the sin(2x).
This equals f'(x).
We are looking for where, again, the tangent lines are horizontal. We are looking for the derivative is equal to
zero. So, we are going to take out this f'(x) and replace it with zero, to find those
horizontal places, where the tangent lines are horizontal. Now, I can take both sides and divide by 2.
Which is just going to cancel that out, and this is going to remain zero.
Now, i've got zero equals cos(x) - sin(2x). Remember solving equations involving trig functions?
Back in trigonometry or pre-calc? You cannot solve for an equation that has two
different trig functions in it, because it acts as two separate variables. It is like having an equation
with an x and a y in it.
And even if they were the same trig function, you have a single angle and a double angle. That is also a problem.
That is also going to act as if there were two separate variables. Normally, when you have one equation,
and two variables, you cannot solve for those two variables. Unless...they factor apart.
That is what is going to happen here. We have zero is equal to cos(x).
Minus...now, all those identities you learned in trig and pre-calc, those three chapters in pre-calc where
it wasn't just a bunch of algebra review. All those identities are still coming back, and they are still
going to be used in Calculus. I need this double angle to come down to a single angle of x.
And the sin(2x) is equal to 2sin(x)cos(x).
Now, I still have one angle - they are all single angles. So that is not an issue now.
But again, I still have those two different trig functions, so I still basically have two variables in this single
equation. That is usually a problem. Except, here I have a cosine function in both terms. If I factor that out, if I
divide it out of those two terms, I get zero is equal to cos(x)...cos(x) divided by cos(x) is one.
And 2sin(x)cos(x) divided by cos(x) is 2sin(x).
My two differnent trig functions... there is only one in the first factor and one in the second.
What do you do after you factor? You set each factor equal to zero, because one of these has to be zero for
the equation to be equal to zero.
When I do that, cos(x) equals zero, and 1-2sin(x) equals zero.
Now I have two equations, and I have one variable in each one of them.
X here is going to be equal to the inverse cosine of zero. And where on the unit circle...within one rotation, zero to
2pi - where is cosine equal to zero? Cosine on the unit circle, again, is x over r. And on the unit circle, because
r is one, it is just x. So where on the unit circle is x the x value equal to zero?
What rotation, which is what you get out of an inverse trig function, will give you a cosine value of zero.
Well, at zero radians, that cosine value is one. So it is at pi over 2...not pi, but 3pi over 2...
So we have pi over 2 and 3pi over 2.
And then we have... I think I made a mistake somewhere...
Yes, only in my notes. Good thing I am redoing this again. Sorry for the interruption.
We have 1-2sin(x)... We are going to bring the one over and get -2sin(x) equals -1.
Divide both sides by -2...and we have sin(x) is equal to one half.
Now we need to get the sine function away from the x. So, x is going to be equal to the inverse sin(1/2).
On the unit circle, sine is y over r. On the unit circle again, we are just looking for the y value.
What rotations on the unit circle have a y value equal to one half?
That is going to be 30 degrees, or pi over 6...and 5pi over 6.
So x is equal to pi over 6 and 5pi over 6.
I am going to go correct my sign error on my notes, while you go do your homework!
I'm Mr. Tarrou!
BAM!
Go do your homework!