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Μάθημα: AP®︎ Λογισμός BC > Ενότητα 5
Μάθημα 3: Determining intervals on which a function is increasing or decreasingFinding decreasing interval given the function
Sal finds the intervals where the function f(x)=x⁶-3x⁵ is decreasing by finding where f' is negative.
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- [Voiceover] So we
have the function f of x is equal to x to the sixth
minus three x to the fifth and we want to know over what
intervals is f decreasing and we're going to do
it without even having to graph y equals f of x and the way we do that is
we look at the derivative of f with respect to x and think about, well, when
is that less than zero? If the rate of change
of f with respect to x is less than zero, well
over those intervals it will be decreasing. So let's first take the derivative. So f prime of x is going to be equal to, just use the power rule here, it's going to be six x to the fifth power minus five times three is 15 x to the, let's decrease that exponent by one, x to the fourth power
and so let's think about when this is going to be less than zero. Over what intervals is six x to the fifth minus 15 x to the fourth less than, less than zero? So let's see, we could factor
out a three x to the fourth, so three x to the fourth times, see if we factor out a
three x to the fourth here, we'll just be left with a two x minus five is less than zero. Did I do that right? Let's see, if I were to distribute it, three times two is six,
x to the fourth times x is x to the fifth, three times five is 15 and x to the fourth, yep, that's right so if I'm getting the
product of two things and I want it to be less than zero, well, there's only way for that to happen is that one or I guess you could say there's two ways for that to happen. Either the first thing is positive and the second is negative or the first is negative
and the second is positive and so let's do that. So either, either three x to the fourth is negative and, and two x minus five is positive or, let me just put the or
in a separate color here, or three x to the fourth is positive and two x minus five is negative. So let's see. This is x to the fourth,
three x to the fourth being less than zero. Well, if we divide both sides by three, this is just going to be x to the fourth needs to be less than zero. And is there any way for something to be to the fourth power to be less than zero? Well, we're assuming we're
dealing with real numbers here and any real number to the fourth power is going to be greater
than or equal to zero so it's actually impossible for something to the fourth
power to be less than zero. This thing is never, this thing is never going
to be less than zero so we can actually rule
out this first case. So we can rule out that
first case right over there. And so we're only going
to worry about this case. So three x to the fourth
being greater than zero, well, that's going to happen as long as x is not equal to zero. So this is, because for any other x,
this is going to be true. x could be negative. You take it to the fourth
power, multiply it by three, it's going to be greater than zero so this is really just the condition that x cannot be equal to zero. And let's see the second one, two x minus five less than zero, that means two x is going
to be less than five and then x is less than five halves. So as long as x is less than five halves and x is not equal to zero, this thing will be decreasing. So if we wanted to write it in terms of, it we wanted to write it
in terms of intervals, we could say x is, let me do this in a new color
just to ease the monotony. So we could say that, having trouble picking a color, so we could say that x is less than zero or zero is less than x is less than five halves. So x is less than zero, this
is all the negative values and then we're essentially
just excluding zero and then going all the way to, all the way to five halves and remember, all I did is say, well, when is our first
derivative negative because if the first
derivative is negative then the rate of change of f
with respect to x is negative or f is decreasing as x is increasing.