Question

me ajudem pf

Seja a função f: R → R tal que f(x) = 6-X/X²

a) Calcule f(-1).

b) Calcule f(3).

c) Calcule f (3/4).

d) Obtenha x tal que f(x) = 1.

e) Obtenha x tal que f(x) = 0.

f) Obtenha x tal que f(x) = -2.​

Answer 1

Resposta:

a) 7

b) 1/3

c) 28/3

d) x₁ = 2, x₂ = -3

e) x = 6

f) x₁ = 1 - √47/4, x₂ = 1-√47/4

Explicação passo-a-passo:

$f\left(x\right)=\frac{6-x}{x^2}$

a) $f_(_-_1_)=\frac{6-\left(-1\right)}{\left(-1\right)^2}$

$\frac{6-\left(-1\right)}{\left(-1\right)^2}\\=\frac{7}{1}\\=7$

b) $f_(_3_)=\frac{6-3}{3^2}$

$\frac{6-3}{3^2}\\=\frac{3}{3^2}\\=\frac{3}{3\cdot \:3}\\=\frac{1}{3}$

c) $f_(_\frac{3}{4} _)=\frac{6-\frac{3}{4}}{\left(\frac{3}{4}\right)^2}$

$\frac{6-\frac{3}{4}}{\left(\frac{3}{4}\right)^2}\\=\frac{6-\frac{3}{4}}{\frac{3^2}{4^2}}\\=\frac{\frac{21}{4}}{\frac{3^2}{4^2}}\\=\frac{21\cdot \:4^2}{4\cdot \:3^2}\\=\frac{3\cdot \:7\cdot \:4}{3^2}\\=\frac{7\cdot \:4}{3}\\=\frac{28}{3}$

d) $1=\frac{6-x}{x^2}$

$1\cdot \:x^2=\frac{6-x}{x^2}x^2\\x^2=6-x\\x_1=2,\:x_2=-3$

e) $0=\frac{6-x}{x^2}$

$0=6-x\\0+x=6-x+x\\x=6$

f) $-2=\frac{6-x}{x^2}$

$-2x^2=\frac{6-x}{x^2}x^2\\-2x^2=6-x\\x_1=\frac{1-\sqrt47}{4}, x_2=\frac{1+\sqrt47}{4}$