If $u (x) = - 2 x^{2} + 3 and v (x) = \frac{1}{x}$ $u(x) = -2x^2+3 and v(x) = 1/x$ , what is the range of $(u \cdot v) (x)$ $(u * v)(x)$ ?

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Answer 1 · 2017-07-23T04:48:59

$u \circ v (x) = u (v (x)) = u (\frac{1}{x}) = \frac{3 x^{2} - 2}{x^{2}}$ $u@v (x) = u(v(x)) = u(1/x) = (3x^2-2)/x^2$

$Range: (- \infty, 3)$ $"Range: " (-oo,3)$

$u (x) = - 2 x^{2} + 3$ $u(x) = -2x^2+3$

$v (x) = \frac{1}{x}$ $v(x) = 1/x$

Whichever notation you prefer:
$(u \circ v) (x) = u (v (x))$ $(u@v) (x) = u(v(x))$

$u \circ v (x) = u (v (x)) = u (\frac{1}{x})$ $u@v (x) = u(v(x)) = u(1/x)$

$u (\frac{1}{x}) = \frac{- 2}{x^{2}} + 3$ $u(1/x) = (-2)/x^2 + 3$

$u (\frac{1}{x}) = \frac{3 x^{2} - 2}{x^{2}}$ $u(1/x) = frac{3x^2-2}{x^2}$

The horizontal asymptote of this function is at $y = 3$ $y=3$ . So, the range of this function is from $(- \infty, 3)$ $(-oo,3)$

If u(x)=−2x2+3andv(x)=1xu(x) = -2x^2+3 and v(x) = 1/x, what is the range of(u⋅v)(x) (u * v)(x)?