I really don't know where to start or go with this question...There aren't any similar examples in my book so if you could walk me through it so I can be prepared for my test - that would be great! Thanks.
(a) Approximate √26 by using the linearization of y=√x at the point (25,5). Show the computation that leads to your conclusion.
(b) Approximate √26 by using a first guess of 5 and one iteration of Newton's method to approximate the zero of x^2-26. Show the computation that leads to your conclusion.
(c) Approximate ³√26 by using an appropriate linearization. Show the computation that leads to your conclusion.
y = m x + b
at (25,5)
5 = m (25) + b
what is m, the slope, there?
y = x^.5
dy/dx = .5 x^-.5 = .5/5 = .1 = m
so
5 = .1 (25) + b
2.5 = b
so our line through (25,5) is
y = .1 x + 2.5
now if x = 26
y = 2.6 + 2.5 = 5.1
that is part a
check it
sqrt 26 = 5.099 Awful close !!! :)
b should come out just like a :)
Thanks so much!! How would I go about C, if you get the chance? A & B make sense now thanks to that walkthrough...I'm just unsure of how to go about it with the cubed root.
Approximate ³√26 by using an appropriate linearization.
y = x^(1/3)
we know that if x = 27, y = 3 so I will start there. Of course you could do something else.
dy/dx = (1/3) x^-2/3
at x = 27
dy/dx = (1/3) /9 = 1/27 = m
so
y = (1/27) x + b
3 = (1/27)(27) + b
b = 2
so
y = (1/27) x + 2
if x = 26
y = 2.963
now for real to check
26^(1/3) = 2.962 We did good again :)
oops! disregard my previous post. thanks so much, I truly appreciate it. Gonna ace this next test :)
Good luck :)
Sure! I can walk you through each part of the question so that you can better understand how to approach it:
(a) To approximate √26 using the linearization of y=√x at the point (25,5), we need to first find the equation of the tangent line to the graph of y=√x at x=25.
The equation of a tangent line to a function at a given point can be found using the point-slope form:
y - y₁ = m(x - x₁), where (x₁, y₁) is the point of tangency and m is the slope of the tangent line.
To find the slope of the tangent line, we differentiate the function y=√x with respect to x:
dy/dx = (1/2) * x^(-1/2)
Substituting x=25 into the derivative, we get:
m = (1/2) * 25^(-1/2) = (1/2) * (1/5) = 1/10
With the slope m = 1/10 and the point (25,5), we can write the equation of the tangent line:
y - 5 = (1/10)(x - 25)
Now, to approximate √26, we can plug in x=26 into the equation of the tangent line and solve for y:
y - 5 = (1/10)(26 - 25)
y - 5 = 1/10
y = 1/10 + 5 = 5.1
Therefore, the approximation for √26 using the linearization is approximately 5.1.
(b) To approximate √26 using Newton's method, we start with an initial guess of 5, which is reasonably close to the actual value of √26.
Newton's method helps us refine an initial guess by iteratively applying the equation:
x₁ = x₀ - f(x₀)/f'(x₀)
where x₀ is our initial guess, f(x) is the function whose zero we want to find (in this case, f(x) = x^2 - 26), and f'(x) is the derivative of f(x) (in this case, f'(x) = 2x).
We start with x₀ = 5 and the equation f(x) = x^2 - 26.
Iteration 1:
x₁ = x₀ - (x₀^2 - 26)/(2x₀)
= 5 - (5^2 - 26)/(2*5)
= 5 - (25 - 26)/10
= 5 - (-1)/10
= 5 + 1/10
= 5.1
Therefore, the approximation for √26 using one iteration of Newton's method is approximately 5.1.
(c) To approximate ³√26 using an appropriate linearization, we can use the linearization of y = ³√x at a convenient point. One such point could be x = 27 (as it is close to 26 and its cube root is a simple value).
To find the equation of the tangent line, we need to find the slope at x = 27.
The derivative of y = ³√x can be found using the chain rule:
dy/dx = (1/3) * x^(-2/3)
Plugging in x = 27, we get:
m = (1/3) * 27^(-2/3) = (1/3) * (1/9) = 1/27
With the slope m = 1/27 and the point (27, ³√27), we can write the equation of the tangent line:
y - ³√27 = (1/27)(x - 27)
Now, to approximate ³√26, we can plug in x = 26 into the equation of the tangent line and solve for y:
y - ³√27 = (1/27)(26 - 27)
y - ³√27 = -1/27
y = -1/27 + ³√27
Therefore, the approximation for ³√26 using the linearization is approximately -1/27 + ³√27.