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3 years ago

X² – 3x – 2 over (x-3)(x - 2)(x - 1)

Mathematics

1 answer:

Lemur [1.5K]3 years ago

5 0

Answer: $\bold{\dfrac{x}{x^2-3x+2}+\dfrac{1}{x^2-3x+2}}$

Step-by-step explanation:

$\dfrac{x^2-3x-2}{(x-3)(x-2)(x-1)} \qquad ;x\neq{1, 2, 3}\\\\\\\\\text{Factor the numerator}\quad \longrightarrow \quad \dfrac{(x-3)(x+1)}{(x-3)(x-2)(x-1)}\\\\\\\text{Cancel out (x-3)}\quad \longrightarrow \quad \dfrac{x+1}{(x-2)(x-1)}\\\\\\\text{To separate them, just place each term in the numerator}\\ \text{over the entire denominator.}\\\\\dfrac{x}{(x-2)(x-1)}+\dfrac{1}{(x-2)(x-1)}\\\\\\=\large\boxed{\dfrac{x}{x^2-3x+2}+\dfrac{1}{x^2-3x+2}}$

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Step-by-step explanation:

Taylor series expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

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$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$