FindING the approximation of a sqroot

Question

A) f(x) = sqr (x) Find the tangent line for x = 25.

 

B) using the tangent line in a, find the approximation of sqr (26)

 

I know how to do part A but I don't know how to do part B. How would you go about doing that?

Answer 1

This looks like a problem where you use the approximation:

 

f(x₀ + Δx) ≅ f(x₀) + f'(x₀)Δx

 

x₀ = 25

 

Δx = 1

 

x₀ + Δx = 5 + 1 = 26

 

f(26) ≅ f(25) + f′(25)(1)

 

f(x) = √x

 

f'(x) = 1 (2√x)

 

f(25) = √25 = 5

 

f'(25) = 1 / (2√25) = 1/10

 

√26 ≅ 5 + (1/10)(1) = 5.1

 

Note that 5.1² = 26.01, which is close.

 

 

Answer 2

Part A gives you a linear equation of the form y = mx + b.  

 

If you evaluate y when x = 26 you will find the approximation of √26.  

Answer 3

A. f(x) = √x   f'(x) = (1/2)x^-1/2 = 1/(2√x)

 

    Slope of tangent line = f'(25) = 1/(2√25) = 1/10

 

    Point on tangent line = (25, f(25)) = (25,5)

 

    Equation of tangent line:  y-5 = (1/10)(x-25)

 

                                           y = (1/10)(x-25) + 5

                                          

B. 

 

    f'(x) = lim_Δx→0[(f(x+Δx) - f(x))/Δx]

 

    So, if Δx ≈ 0, f'(x) ≈ (f(x+Δx) - f(x))/Δx

 

                       f'(x)Δx ≈ f(x+Δx) - f(x)

 

                        f(x+Δx) ≈ f'(x)Δx + f(x)  (**)

 

                         If we let x = 25, x+Δx = 26 (so Δx = 1), f(x) = √x,                                and f'(x) = 1/(2√x), then (**) becomes:

 

                         √26 ≈ (1/10)(1) + 5 = 5.1 

 

                           This is what we get for y by plugging in 26 for x into

                           the tangent line equation y = (1/10)(x-25)+5.   

 

Answer 4

f'(x) = ½(x^-½)

f'(25) = 1/10 = Δy ÷ Δx

x = 25, Δx = 1; Δy = 0.1, 26 = 25 + Δx

f(26) ≈ f(x) + Δy = f(25) + 0.1 = 5.1

 

This is a pretty good approximation because square root of 26 is 5.099019514 on my little calculator.