Video Transcript
By finding the linear approximation
of the function 𝑓 of 𝑥 is equal to the square root of 𝑥 as a suitable value of
𝑥, estimate the value of the square root of 100.5.
The question is asking us to
estimate the value of the square root of 100.5 by finding a linear approximation of
the function 𝑓 of 𝑥 is equal to the square root of 𝑥 at a suitable value for
𝑥. Let’s start by recalling what a
linear approximation of a function 𝑓 of 𝑥 at the point 𝑥 equals 𝑎 is. If 𝑓 is differentiable at 𝑥 is
equal to 𝑎, then we can approximate our function 𝑓 of 𝑥 near 𝑥 is equal to 𝑎 by
using the tangent line. We call this a linear
approximation, 𝐿 of 𝑥. It’s equal to 𝑓 evaluated at 𝑎
plus the first derivative of 𝑓 of 𝑎 times 𝑥 minus 𝑎. So how are we going to use this to
estimate our value of the square root of 100.5?
Since we’re asked to do this by
finding a linear approximation of the square root of 𝑥, we need to choose our value
of 𝑎. Remember, our approximation will be
more accurate the closer we are to 𝑥 is equal to 𝑎. So we should choose a value of 𝑎
which gives us a value close to the square root of 100.5. If we choose 𝑎 is equal to 100,
then we get that 𝑓 evaluated at 𝑎 is the square root of 100. And this is close to the square
root of 100.5. So we’ll take this value of 𝑎. So to find our linear
approximation, we need 𝑓 evaluated at 𝑎 and 𝑓 prime evaluated at 𝑎. We already found 𝑓 evaluated at
𝑎; it’s the square root of 100. So let’s now find the first
derivative of 𝑓 evaluated at 𝑎.
We need to differentiate 𝑓 of 𝑥
is equal to root 𝑥. To do this, we’ll start by
rewriting 𝑓 of 𝑥 by using our laws of exponents. Root 𝑥 is equal to 𝑥 to the power
of one-half. We can then differentiate this by
using the power rule for differentiation. We multiply by our exponent of 𝑥
and reduce this exponent by one. This gives us 𝑓 prime of 𝑥 is
equal to one-half times 𝑥 to the power of negative one-half. And again, we will rewrite this by
using our laws of exponents. 𝑥 to the power of negative
one-half is the same as dividing by the square root of 𝑥. So 𝑓 prime of 𝑥 is equal to one
divided by two root 𝑥.
We’re now ready to find 𝑓 prime of
𝑎. Since 𝑎 is equal to 100, this is
the first derivative of 𝑓 at 𝑥 is equal to 100. So we substitute 𝑥 is equal to 100
into our expression for 𝑓 prime of 𝑥. We get one divided by two root
100. And the square roots of 100 is
10. So 𝑓 prime of 100 is equal to one
divided by 20. So we’re now ready to find the
linear approximation of our function 𝑓 of 𝑥 is equal to the square root of 𝑥 at
100. We get 𝐿 of 𝑥 is equal to 𝑓 of
100 plus 𝑓 prime of 100 times 𝑥 minus 100. We know that 𝑓 of 100 is the
square root of 100 and 𝑓 prime of 100 is one divided by 20. So 𝐿 of 𝑥 is equal to root 100
plus one over 20 times 𝑥 minus 100.
However, we can simplify this
further. The square root of 100 is equal to
10. Next, we can distribute one
twentieth over our parentheses. We get 𝑥 over 20 minus 100 over
20. And 100 over 20 is equal to
five. Finally, 10 minus five is equal to
five. So our linear approximation of the
square root of 𝑥 at 100 is equal to five plus 𝑥 over 20. We’re now ready to approximate the
square root of 100.5 by using our linear approximation. The square root of 100.5 is
approximately equal to 𝐿 evaluated at 100.5. Substituting 𝑥 is equal to 100.5
into our expression for 𝐿 of 𝑥, we get five plus 100.5 divided by 20, which if we
calculate, we get 10.025.
Therefore, by finding the linear
approximation of the function 𝑓 of 𝑥 is equal to the square root of 𝑥 at 𝑥 is
equal to 100, we’ve shown the square root of 100.5 is approximately equal to
10.025.
