Home > Calculus > Using the Tangent Line to Approximate Function Values

Use Newton's method to approximate the given number correct to eight decimal places?

95^(1/95)

Answer 1

Scarlett Allison

#1.04910303 (8 d.p.)#

Let #x=95^(1/95)#

#x^95=95#

#x^95-95=0#

Let #f(x)=x^95-95#

Then #f'(x)=95x^94#

Now we simply plug into the formula with a starting point.

#x_0=1.2#

#x_1=x_0-(f(x_0))/(f'(x_0))#

#=1.2-(1.2^95-95)/(95(1.2)^94)#

#=1.18736845#

Second iteration:

#x_2=x_1-(f(x_1))/(f'(x_1))#

#=1.17486993#

And so on, starting with #1.2# it took me 15 iterations to get to an approximation with 8 constant d.p. Answer should be #1.04910303 (8 d.p.)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezekiel Alexander

Sure, I can help with that. Could you please provide the specific number you'd like to approximate using Newton's method?

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

Ezekiel Alexander

To approximate a number using Newton's method, you typically need an initial guess ( x_0 ) and a function ( f(x) ). Then you iterate using the formula:

[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} ]

You continue iterating until the value of ( x ) stabilizes. You can use this method to approximate square roots or zeroes of functions, for example. If you have a specific number and function in mind, please provide them, and I can demonstrate the process for you.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.