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

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

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

taylor

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

5 + 4 x - 3 x^{2} + x^{3}

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 5 + \frac{4}{1 !} x + \frac{- 6}{2 !} x^{2} + \frac{6}{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

= 5 + 4 x - 3 x^{2} + x^{3}

(1 + x) cos (x), at x = 0

9 x - x^{2}, 2 x

x (x^{2} + 9)

\frac{\partial}{\partial x} (sin (x e^{x^{2} y}))

\frac{\partial}{\partial x} (ln (x^{4} + y^{2}) - z)

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