Linear Approximation

Question

Use linear approximation or the tangent line to approximate 1/ 0.254 as follows:

Let f(x) = 1/x and find the equation of the tangent line to f(x) at nice points near 0.254. Then use this to approximate 1/0.254.

Answer 1

To solve a linear approximation problem like this, the first thing to do is to identify a nice point near the given one. This should be a number a for which f(a) is easy to calculate. A nice number near 0.254 is 0.25 = 1/4, because it's easy to figure out that f(0.25) = 1/(1/4) = 4.

Next, we need to find the derivative of the function. The function is f(x) = 1/x = x^{-1}, so using the power rule,

the derivative is

f'(x) = -x^{-2} = -1/(x^2).

Plugging in a=0.25, we get f'(0.25) = -1/(0.25^2) = -16.

Now we can put everything together. The tangent line is the line whose slope is f'(a) and which passes through the point (a, f(a)). In our case, a=0.25, f(a)=4, and f'(a)=-16, so we need a line with slope -16 passing through (0.25, 4). Using the point-slope form, the equation is

y-4=-16(x-0.25).

Finally, we use the tangent line equation to make our approximation. This means that we plug in the actual value for x into the equation of the tangent line and solve for y. So plug in x=0.254 to the equation above:

y-4 = -16 (0.254-0.25)

y-4 = -16(0.004)

y-4 = -0.064

y = 4- 0.054 = 3.936

So we conclude that 1/0.254, which equals f(0.254) is approximately 3.936.

Answer 2

To approximate 1/0.254 using linear approximation, we'll use the function

f(x)=1/x and find the equation of the tangent line at a point x_0​ close to 0.254.

Step 1: Choose a "nice" point x0​ near 0.254

A "nice" point close to 0.254 is x0=0.25, because it's a simple fraction and easier to work with.

Step 2: Find the derivative of f(x)=1/x

f′(x)=−1/x^2

Step 3: Find the slope of the tangent line at x0=0.25

i think you can complete the rest