Question

найти область значений функции f(x)=x²-4x+8/x²-4x+5.
pls))

Answer 1

Ответ:

$f(x) = {x}^{2} - 4x + \frac{8}{ {x}^{2} } - 4x + 5 \\ f (x) = \frac{d}{dx} ( {x}^{2} - 4x + \frac{8}{ {x}^{2} } - 4x + 5) \\ f(x) = \frac{d}{dx} ( {x}^{2} - 8x + \frac{8}{ {x}^{2} } + 5) \\ f(x) = \frac{d}{dx} ( {x}^{2} ) + ( - 8x) + \frac{d}{dx} ( \frac{8}{ {x}^{2} } ) + \frac{d}{dx} (5) \\ f(x) = 2x - 8 - 8 \times \frac{2x}{( {x}^{2})^{2} } + 0 \\ f(x) = 2x - 8 - \frac{16}{ {x}^{3} }$