Approximation problem HELP PLEASE!!!

[HousEDM]

Use linear approximation, i.e. the tangent line, to approximate (1)/(0,251) as follows: Let f(x) = 1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.251. Then use this to approximate (1)/(0.251)

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i dont understand how this is the wrong answer help would be really appreciated.

 

[Ishuda]

Use linear approximation, i.e. the tangent line, to approximate (1)/(0,251) as follows: Let f(x) = 1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.251. Then use this to approximate (1)/(0.251)

<link removed> i dont understand how this is the wrong answer help would be really appreciated.

A 'nice point' would probably be 0.25 since that is 1/0.25 = 1/(1/4) = 4 easily computed. For the tangent line you need the derivative, then use the point-slope form to write the tangent line. The tangent line should be close to the curve at x = 0.251, so just evaluate the the tangent line there for an estimate of 1/.251.

 

[HousEDM]

A 'nice point' would probably be 0.25 since that is 1/0.25 = 1/(1/4) = 4 easily computed. For the tangent line you need the derivative, then use the point-slope form to write the tangent line. The tangent line should be close to the curve at x = 0.251, so just evaluate the the tangent line there for an estimate of 1/.251.

yeah thats what i tried, the tangent line with nice point 0.25 is y=-1/16x + 1/2 but when plugging in x as 1/.251 it says its incorrect

 

[Deleted member 4993]

yeah thats what i tried, the tangent line with nice point 0.25 is y=-1/16x + 1/2 but when plugging in x as 1/.251 it says its incorrect

Where did you plug in what?

Please show complete work!

 

[HousEDM]

to find the slope of the tangent i found the derivative of 1/x which is -1/x^2 and plugged in my "nice" point to find M, slope of the tangent, there -1/(1/.25)^2 = -1/16
then found b of the tangent line by using Point(.25,1/.25)

then the equation of the tangent line is -1/16x + .5

then they ask to find linear approximation of 1/.251 and i don't know whether you plug in 1/.251 or just .251

 

[Deleted member 4993]

m = y'(x₀) = [y(x₁) - y(x₀)]/(x₁ - x₀)

x₁ = 0.251

x₀ = 0.25

y(x₀) = 1/(0.25) = 4

y'(x₀) = -16 .........................edited

Now calculate y(x₁)

 

[Ishuda]

yeah thats what i tried, the tangent line with nice point 0.25 is y=-1/16x + 1/2 but when plugging in x as 1/.251 it says its incorrect

f(x) = 1/x
f(.25)= 4
f'(x) = -1/x²
f'(.25) = - 16
Tangent line, slope m = -16 and point (x₀,y₀)= (0.25, 4). Point - slope form of line
y = y₀ + m (x - x₀)

 

[HousEDM]

Thanks guys it worked