# Given #f(x)=root3 (1+3x)# at a=0 and use it to estimate the value of the #root3( 1.03)#?

There are a couple of approaches to get a really good approximation

Method 1 - Binomial Series (Power Series) Expansion:

The first is to use the Binomial series expansion, which is essentially the Taylor Series about

So using the Binomial Series we have;

# f(x) = root(3)(1+3x) #

# " " = (1+3x)^(1/3) #

# " " = 1 +(1/3)(3x) + (1/3)(-2/3)(3x)^2/(2!) + #

# " "(1/3)(-2/3)(-5/3)(3x)^3/(3!) + ...#

# " " = 1 + x -x^2+5/3x^3 + ... #

And if we put

# f(0.01) = root(3)(1+0.03) #

# " " = root(3)(1.03) #

# " " = 1 + 0.01 -0.01^2+5/3(0.01)^3 + ... #

# " " = 1 + 0.01 -0.0001+5/3(0.000001) + ... #

# " " ~~ 1 + 0.01 -0.0001+0.000001667 # (ignoring higher terms)

# " " ~~ 1.009901667 #

And so it would be reasonable to conclude that

And in fact using a calculator we find

Method 2 - Newton-Rhapson

Another approach is to use Newton Rhapson to solve an equation of which

# \ \ \ \ \ \ \x = root(3)(1.03) #

# :. x^3 = 1.03 #

# :. x^3 - 1.03 = 0 #

Let

First let us look at the graphs:

graph{x^3 - 1.03 [-5, 5, -15, 15]}

We can see there is one solution in the interval

We can find the solution numerically, using Newton-Rhapson method

# \ \ \ \ \ \ \f(x) = x^3 - 1.03 #

# :. f'(x) = 3x^2 #

The Newton-Rhapson method uses the following iterative sequence

# { (x_0,=1), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :} #

Then using excel working to 8dp we can tabulate the iterations as follows:

We could equally use a modern scientific graphing calculator as most new calculators have an " *Ans* " button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution is

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To estimate the value of ( \sqrt[3]{1.03} ) using the given function ( f(x) = \sqrt[3]{1+3x} ) at ( a=0 ), we can use linear approximation (the tangent line at ( a=0 )). The linear approximation is given by:

[ f(x) \approx f(a) + f'(a)(x-a) ]

where ( f'(x) ) denotes the derivative of ( f(x) ) with respect to ( x ).

First, find the derivative of ( f(x) ) and evaluate it at ( a=0 ). Then, plug in ( a=0 ) and ( x=0.03 ) into the linear approximation formula to estimate the value of ( \sqrt[3]{1.03} ).

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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.

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.

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.

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