Use linear approximation, i.e. the tangent line, to approximate (1/0.504) as follows: Find the equation of the tangent line to f(x)=1/x at a "nice" point near 0.504. Then use this to approximate (1/0.504).

The idea behind the approximation is:

y(x)=1/x
y'(x)=-1/x²
Let x=x0+Δx
y(x) = y(x0)+y'(x0)*Δx
=y(x)-Δx / (x0 ²)

Now apply x0=0.5, Δx=0.004

y(0.504)=y(0.5)-0.004/0.5²
=2-0.016
=1.984
compare to approximate actual value of
1.984126984126984

To approximate (1/0.504) using linear approximation, we need to find the equation of the tangent line to the function f(x) = 1/x at a point near 0.504.

First, let's find the derivative of f(x). The derivative of f(x) = 1/x can be found using the power rule for differentiation:

f'(x) = -1/x^2

Now, let's find a "nice" point near 0.504. Let's choose x = 0.5 as our point.

Using this point, we can find the slope of the tangent line. The slope of the tangent line is equal to the value of the derivative at the chosen point.

f'(0.5) = -1/(0.5)^2 = -1/0.25 = -4

The slope of the tangent line is -4.

Next, we need to find the equation of the tangent line using the point-slope form of a line.

The equation of the tangent line is given by:

y - y₁ = m(x - x₁)

where (x₁, y₁) is the point on the line, m is the slope of the line.

Using (x₁, y₁) = (0.5, 2) and m = -4, we have:

y - 2 = -4(x - 0.5)

Simplifying, we get:

y = -4x + 4

Now, we can use this equation to approximate (1/0.504). Since 0.504 is close to 0.5, we can substitute 0.504 for x in the equation:

y = -4(0.504) + 4

Simplifying, we find:

y ≈ 1.984

Therefore, using linear approximation, (1/0.504) is approximately equal to 1.984.

To approximate (1/0.504) using linear approximation, we first need to find the equation of the tangent line to the function f(x) = 1/x at a nice point near 0.504.

Step 1: Find the derivative of f(x)
The derivative of f(x) = 1/x can be found using the power rule for differentiation.
f'(x) = -1/x^2

Step 2: Find the slope of the tangent line
We need to find the slope of the tangent line at a nice point near 0.504. To do this, we substitute the value of x into the derivative.
Let's consider a point near 0.504, for example, x = 0.5. Substituting x = 0.5 into f'(x) = -1/x^2, we get:
f'(0.5) = -1/(0.5)^2 = -1/0.25 = -4

Step 3: Find the equation of the tangent line
We now have the slope of the tangent line, which is -4. To find the equation of the tangent line, we need to determine the value of f(x) at the chosen point (0.5). Since f(x) = 1/x, when x = 0.5, f(x) = 1/0.5 = 2.

The equation of the tangent line in point-slope form is given by:
y - y1 = m(x - x1)

Substituting the values, we have:
y - 2 = -4(x - 0.5)

Simplifying, we get:
y - 2 = -4x + 2

Step 4: Use the tangent line to approximate (1/0.504)
To approximate (1/0.504) using the tangent line, we substitute x = 0.504 into the equation of the tangent line.

y - 2 = -4(0.504) + 2

Simplifying, we get:
y - 2 = -2.016 + 2

y - 2 ≈ -0.016

Finally, we solve for y to find the approximation:
y ≈ 1.984

Therefore, the linear approximation to (1/0.504) using the tangent line is approximately 1.984.