Here’s the transcription formatted for closed captions:
We want to use a tangent line to approximate the square root of forty-nine point one. To do this,
we'll be using the square root function f of x equals the square root of x and we'll use
the tangent line at x equals forty-nine. We're using x equals forty-nine for two reasons. First,
it's close to forty-nine point one and second, the square root of forty-nine
simplifies perfectly. So the idea is that we can use the equation of this red tangent line at x
equals forty-nine to make an approximation for the green function at x equals forty-nine point one.
Notice how the tangent line is a good approximation for the function
as long as x is close to forty-nine. But as we get further and further away from x equals forty-nine,
the approximations would have more of an error. So to find the equation of the tangent line,
we'll have to find the point of tangency as well as the slope of the tangent line. Well,
we know x equals forty-nine so that would be the x coordinate of the point of tangency.
And because the point of tangency is on the square root function,
we can evaluate f of forty-nine to find the corresponding y coordinate. So f of forty-nine
would just be the square root of forty-nine which is equal to seven. So the point of
tangency is the point forty-nine comma seven. And now to find the slope of the tangent line,
we can find the derivative function and evaluate the derivative function at x equals forty-nine.
So let's work on finding our derivative function. If f of x equals the square root of x because the
index is two and the exponent on x is one, we can write this as x to the one-half. And now we can
find f prime of x by applying the power rule of differentiation. f prime of x would be equal to
one-half times x to the power of one-half minus one. That would be negative one-half.
To change the sign of the exponent, we can move this to the denominator and write this as
one divided by two x to the one-half. And x to the one-half is equal to the
square root of x. So we can write this as one divided by two square root x,
which means the slope of our tangent line at x equals forty-nine would be equal to f
prime of forty-nine which would be equal to one divided by two square root forty-nine.
Square root forty-nine is equal to seven so we have one divided by
two times seven or one fourteenth. So the slope of the tangent line is one fourteenth.
And now we have the information we need to find the equation of the tangent line.
So let's continue on the next slide using point-slope form of a line
where x sub one is equal to forty-nine and y sub one is equal to seven.
We would have y minus seven equals one fourteenth times the quantity x minus forty-nine.
So now we would distribute so we'd have y minus seven equals one fourteenth x and then we'd have
minus one fourteenth times forty-nine or times forty-nine over one. There are two sevens in seven
and seven sevens in forty-nine so we'd have minus seven halves. Add seven to both sides so we have
y equals one fourteenth x minus seven halves plus seven over one. Common denominator would be two,
multiply seven over one by two over two. So our linear equation would be y equals one fourteenth x
and then here we'd have negative seven halves plus fourteen halves, that's positive seven halves.
This would be the equation of our tangent line which we'll use to make our approximation for
the square root of forty-nine point one. Let's go ahead and call this l of x equals
one fourteenth x plus seven halves.
So now to make our approximation, we'll find l of forty-nine point one. So we'll have one fourteenth
times because we want the exact value of this forty-nine point one
is equal to forty-nine and one tenth.
If we convert this to an improper fraction, we'd have tenths
and the numerator would be ten times forty-nine plus one or four hundred ninety-one tenths. So
we'll go ahead and use this value here four hundred ninety-one tenths and then
plus seven halves. Notice our denominator here is going to be one hundred forty so to write seven
halves as one hundred fortieths, we'll have to multiply the numerator and denominator by seventy.
So we'd have four hundred ninety-one one hundred fortieths plus four hundred ninety
one hundred fortieths. Now that we have a common denominator, our approximation for
the square root of forty-nine point one would be nine hundred eighty-one one hundred fortieths
which does not simplify. So this would be our exact approximation using our tangent line.
But if we were asked to round this to four decimal places, let's go ahead and do that as well.
Nine hundred eighty-one divided by one hundred forty
to four decimal places would be approximately seven point zero zero seven one.
It's always nice to get the decimal approximation because it does give us an idea
of whether our estimate is appropriate. Notice how because the square root of forty-nine is seven,
the square root of forty-nine point one should be a little larger than seven, which it is.
I hope you found this helpful.