Question

va rog repedeeee......​

Answer 1

a)

$f(m)=2\iff 2m+3=2 \iff 2m = -1 \implies m = -\dfrac{1}{2}$

b)

$f(-1) = \dfrac{1}{2} \iff -m(-1)+m=\dfrac{1}{2} \iff m+m = \dfrac{1}{2}$

$2m=\dfrac{1}{2} \implies m = \dfrac{1}{4}$

c)

$f(m) = m \iff -m+m = m \iff 0 = m \implies m = 0$

d)

$f\Big(\dfrac{1}{2}\Big) = m \iff \dfrac{m}{3} \cdot \dfrac{1}{2} - 1 = m \iff \dfrac{m}{6} - 1 = m$

$m - 6 = 6m \iff 6m-m=-6 \iff 5m=-6 \implies m = -\dfrac{6}{5}$

e)

$f(2m) = - m \iff -3 \cdot 2m + 5m = -m \iff -6m + 5m + m = 0$

$6m-6m=0 \iff 0\cdot m = 0 \implies m \in \mathbb{R}$

f)

$f(-1) = \sqrt{2} \iff 3 \cdot (-1) - \dfrac{m}{\sqrt{2} } = \sqrt{2}$

$-3 - \dfrac{m}{\sqrt{2} } = \sqrt{2} \iff - \dfrac{m}{\sqrt{2} } = 3 + \sqrt{2}\ \ \Big|\cdot (-\sqrt{2})$

$m = -\sqrt{2} (3 + \sqrt{2}) \implies m = -3\sqrt{2} - 2$