Tempo atual:0:00Duração total:6:24

0 pontos de energia

Studying for a test? Prepare with these 2 lessons on Existence theorems.

See 2 lessons

# Exemplo do teorema de Lagrange: função de raiz quadrada

Transcrição de vídeo

- [Voiceover] Let f of x
be equal to the square root of four x minus three,
and let c be the number that satisfies the mean
value theorem for f on the closed interval
between one and three, or one is less than or equal to x is less than or equal to three. What is c? So let's just remind ourselves
what it means for c to be the number that satisfies
the mean value theorem for f. This means that over this
interval, c is a point, x equals c is a point where
the slope of the tangent line at x equals c, so I
could write f prime of c, so that is the slope of the tangent line when x is equal to c, this is equal to the slope of the secant line that connects these two points. So this is going to be equal to, see the slope of the
secant line that connects the points three, f of
three, and one, f of one. So it's going to be f
of three minus f of one over three minus one. And if you wanted to think
about what this means visually, it would look something like this. So if this is our x axis and this is one, two, actually let me spread
it out a little bit more. One, two, and three. And so you have one, comma,
f of one, right over there, so that is at the point
one, comma, f of one, and we could evaluate that, actually, that's one, comma, one. So that's gonna be the
point one, comma, one. And then you have the point three, comma, let's see, you're gonna
have four times three is 12 minus three is nine, so it's
gonna be three, comma, three. So maybe it's right over
there, three, comma, three, and the curve might look
something like this, so it might look something like that, so if you think about
the slope of the line that connects these two points, so this line that
connects those two points, all the mean value theorem, I'm doing a different color. All the mean value
theorem tells us is that there's a point between
one and three where the slope of the tangent line
has the exact same slope. So if I were to eyeball it, it looks like it's right around there, although we are actually
going to solve for it. So, some point where the
slope of the tangent line is equal to the slope of the line that connects these two end points and their corresponding function values. So that is c, that would
be c right over there. So really we just have to solve this. So let's first just find
out what f prime of x is, and then we could substitute a c in there and then we can evaluate
this on the right hand side. So I'm gonna rewrite f of x, f of x is equal to, and I'm gonna write it as four x to the minus
three to the one half power. Makes it a little bit more
obvious that we can apply the power rule and the chain rule here. So f prime of x is going
to be the derivative of four x minus three to the one half with respect to four x minus three, so that is going to be
one half times four x minus three to the negative one half, and then we're gonna multiply that times the derivative of four x minus three with respect to x. Well, derivative of four x
with respect to x is just four, and the derivative of negative
three with respect to x, well that's just gonna be zero, so the derivative of four x minus three with respect to x is four. So times four. so f prime of x is equal to four times one half, which is two, over the square
root of four x minus three. Four x minus three to the one half would just be the square
root of four x minus three, but it's the negative one half, so we're gonna put it in the
denominator right over here. And so f prime of c, we could rewrite this as two over the square
root of four c minus three, and what is that going to be equal to? That is going to be equal to, let's see, f of three we already
figured out is three, f of one we already figured out is one, and so we get three minus
one over three minus one, well that's gonna be two over
two, which is equal to one. So there's some point
between one and three where the derivative at that point, the slope of the tangent
line, is equal to one. So let's see if we can solve
this thing right over here. We can multiply both
sides of this by four c by the square root of four c minus three, and so then we are going to get two is equal to the square
root of four c minus three. All I did is multiply both sides of this by square root of four c minus three to get rid of this in the denominator, and so let's see, now we
get rid of the radical, we can square both sides, and so actually let me just show that, so now we can square both sides, so we get four is equal
to four c minus three, add three to both sides,
seven is equal to four c, and then divide both sides by four. I'll go right here to do it, you're going to get c is
equal to seven fourths. C is equal to seven over four, which is equal to one and three fourths, or we could view this as 1.75. So actually, the c value
is a little bit closer, I hand drew this, it's closer
to about right over there on our diagram, actually
that looks pretty close, that actually looks pretty good, I just hand drew this curve
so it's definitely not exact. But anyway, hopefully that gives you a sense of what's going on here. We're just saying, hey,
the mean value theorem gives us some c where the
slope of the tangent line is the same as the slope of the line that connects one, f of
one, and three, f of three.