# How do you find the smallest value of #n# for which the Taylor series approximates the function #f(x)=e^(2x)# at #c=2# on the interval #0<=x<=1# with an error less than #10^(-6)#?

Upon determining the remainder term,

Since

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To find the smallest value of ( n ) for which the Taylor series approximates the function ( f(x) = e^{2x} ) at ( c = 2 ) on the interval ( 0 \leq x \leq 1 ) with an error less than ( 10^{-6} ), we use the remainder term formula for Taylor series.

The Taylor series expansion for ( f(x) = e^{2x} ) centered at ( c = 2 ) is:

[ f(x) \approx \sum_{k=0}^{n} \frac{f^{(k)}(2)}{k!} (x - 2)^k ]

The remainder term is given by:

[ R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!} (x - 2)^{n+1} ]

where ( z ) lies between ( 2 ) and ( x ).

We want ( R_n(x) < 10^{-6} ) for ( 0 \leq x \leq 1 ). So, we need to find the maximum value of ( f^{(n+1)}(z) ) on the interval ( 0 \leq x \leq 1 ).

Since ( f(x) = e^{2x} ), ( f^{(n+1)}(x) = 2^{n+1}e^{2x} ), which is maximized at ( x = 1 ).

So, ( f^{(n+1)}(z) \leq 2^{n+1}e^{2} ) for ( 0 \leq z \leq 1 ).

Now, we solve for ( n ) such that:

[ \frac{2^{n+1}e^{2}}{(n+1)!} (1 - 2)^{n+1} < 10^{-6} ]

[ \frac{2^{n+1}e^{2}}{(n+1)!} < 10^{-6} ]

We can use numerical methods or approximation techniques to solve for the smallest ( n ).

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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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