# How do you find the first five terms of the Taylor series for #f(x)=x^8+x^4+3# at #x=1#?

or, with the series notation:

So, let's calculate all the derivatives we need:

so:

It's easy to understand that if you would calculate the Taylor's series until the 8th order you will obtain the function itself, because it is a polynomial!

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To find the first five terms of the Taylor series for ( f(x) = x^8 + x^4 + 3 ) at ( x = 1 ), we need to find the derivatives of ( f(x) ) and evaluate them at ( x = 1 ). Then, we'll use these values to write out the Taylor series expansion.

First, let's find the derivatives of ( f(x) ):

( f'(x) = 8x^7 + 4x^3 )

( f''(x) = 56x^6 + 12x^2 )

( f'''(x) = 336x^5 + 24x )

( f''''(x) = 1680x^4 + 24 )

( f'''''(x) = 6720x^3 )

Now, let's evaluate these derivatives at ( x = 1 ):

( f(1) = 1^8 + 1^4 + 3 = 5 )

( f'(1) = 8(1)^7 + 4(1)^3 = 8 + 4 = 12 )

( f''(1) = 56(1)^6 + 12(1)^2 = 56 + 12 = 68 )

( f'''(1) = 336(1)^5 + 24(1) = 336 + 24 = 360 )

( f''''(1) = 1680(1)^4 + 24 = 1680 + 24 = 1704 )

Now, let's write out the first five terms of the Taylor series expansion:

( f(x) \approx f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + \frac{f''''(1)}{4!}(x-1)^4 )

( = 5 + 12(x-1) + \frac{68}{2!}(x-1)^2 + \frac{360}{3!}(x-1)^3 + \frac{1704}{4!}(x-1)^4 )

Simplify if needed.

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