We are given f of one equals two
and f prime of one equals negative two.
We're asked to estimate the following function values:
f of two, f of 1.5, and f of 1.25.
From the given information, we will need to use tangent
line approximations to approximate these function values.
I provided a graph below to help explain the
process for approximating these function values.
For a quick review, the idea is we can use
a tangent line to approximate function values
as long as we are near the point of tangency.
Notice how on this graph, the red line is
tangent to the blue function at x equals one.
If we zoom in on the point of tangency and use
the red tangent line to approximate, let's say,
this function value here, the idea is the change
of y along the tangent line, given by differential y,
is approximately equal to the change of y along the
original function given by this vertical distance delta y.
And we can say that differential y is approximately
equal to delta y as long as we are near the point
of tangency, which is why we can use a tangent
line approximation to approximate function values.
Now looking at the horizontal change from this first
point to the second point, notice how differential x,
the change of x along the tangent line, is equal to
delta x, the change of x along the original function.
Now let's go back and make our tangent line approximations.
To approximate f of two, we will begin with
the function value x equals one, and then
add the change in y along the tangent line,
and therefore f of two is approximately
equal to f of one plus the change of y
along the tangent line, is differential y.
Remember, differential y is equal to f prime of x times dx,
and therefore this is equal to f of one plus, differential
y is equal to f prime of one times differential x.
We know f of one is two, and f prime of one is equal
to negative two, let's determine differential x.
Looking at the graph below, we're beginning at the point
of tangency where x equals one, and then we're trying
to approximate the function value when x equals two.
If we move from x equals one to x equals two,
x is increasing by one, and therefore delta x
equals differential x equals positive one.
Going back to our calculation, this would give us
f of one, which equals two, plus f prime of one, which
is negative two, times differential x, which is one, this
gives us two plus negative two, which is equal to zero, our
tangent line approximation for f of two is equal to zero.
If we go back to the graph, looking at the red tangent line,
notice how when x equals two, the function value is zero.
Looking at the original function, though, notice how
f of two is x equal to negative one, this difference of
one unit is the error by using this linear approximation.
Now to approximate f of 1.5.
We'll begin by using the function value x equals one,
which is f of one, and then add the change of y along
the tangent line, which is differential y, which is equal
to f of one plus, differential y equals again f prime of one
times differential x, but now differential x is different
because we're going form x equals one, the point of tangency
to x equals 1.5, which would be approximately here.
So because x is increasing by 0.5, delta x,
which equals differential x, now is equal to 0.5.
Going back to the calculation, this gives us two,
which is f of one, plus f prime of one,
which is negative two, times differential x,
which is 0.5, negative two times 0.5 is negative one,
two plus negative one is equal to positive one.
Our tangent line approximation for f of 1.5 is positive one.
If we look at the red tangent line, notice how when x
equals 1.5, the function value for the tangent line
is positive one, but the true function value is
less than one, approximately here, but notice how
for this approximation, the error is very small.
And now let's look at one last approximation.
We want to approximate f of 1.25, which is approximately
equal to the function value x equals one, or f of one,
plus the change of y along the tangent line, which is
equal to f of one plus f prime of one times differential x.
Let's determine differential x.
We begin at x equals one and move to x
equals 1.25, which is approximately here.
Notice how we're increasing x by 0.25 units,
and therefore delta x or differential x equals 0.25.
Going back to our calculation, this gives us
two plus negative two times 0.25 or one-fourth,
which gives us two plus, negative two times 0.25 is equal
to negative one-half or 0.5, which is equal to 1.5.
If we go back to the graph one last time, if x equals 1.25,
notice how it's almost impossible to determine the
difference between the function value given by
the tangent line and the true function value
of the blue function, which means the error is
very, very small when approximating f of 1.25.
I hope you found this helpful.