What is taylor f(x)=3^3-3(3)^2+4(3)+5+x ?

The answer to taylor f(x)=3^3-3(3)^2+4(3)+5+x is 17+x

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taylor f(x)=3^3-3(3)^2+4(3)+5+x

taylor

f (x) = 3^{3} - 3 (3)^{2} + 4 (3) + 5 + x

17 + x

Solution steps

Apply the Taylor Formula

= 17 + \frac{\frac{d}{d x} ( 3 ^{3} - 3 \cdot 3 ^{2} + 4 \cdot 3 + 5 + x ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 3 ^{3} - 3 \cdot 3 ^{2} + 4 \cdot 3 + 5 + x ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 3 ^{3} - 3 \cdot 3 ^{2} + 4 \cdot 3 + 5 + x ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 17 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{0}{10 !} x^{10}

Refine

= 17 + x

What is taylor f(x)=3^3-3(3)^2+4(3)+5+x ?
The answer to taylor f(x)=3^3-3(3)^2+4(3)+5+x is 17+x