Ada yang bisa bantu jawab soal fungsi invers? :(.... Question from @feisa21

Gusti2601

$\mathbb \color{aqua} \underbrace{JAWABAN}$

SOAL 1 :

$\boxed{ \small \tt {f}^{ - 1} (x) = \bf \dfrac{x - 4}{2}}$

$\small \boxed{ \tt {f}^{ - 1} (2) = \bf - 1} \\$

SOAL 2 :

$\boxed{ \small \tt {f}^{ - 1} (x) = \bf \dfrac{ - 3x - 7}{7x - 4} } \\$

SOAL 3 :

$\boxed{ \small \tt {f}^{ - 1} (x) = \bf 3 \pm \sqrt{x - 6} }$

$\boxed{ \small \tt {f}^{ - 1} (1) = \bf 3 + \sqrt{ - 5} }$

$\boxed{ \small \tt {f}^{ - 1} (1) = \bf 3 - \sqrt{ - 5} } \\$

SOAL 4 :

$\boxed{ \small \tt {g}^{ - 1} (x) = \bf \frac{ - 6x - 1}{ - 2} } \\$

SOAL 5 :

$\boxed{ \small \tt {f}^{ - 1} (x) = \bf x + 3 }\\$

SOAL 6 :

$\boxed{ \small \tt {f}^{ - 1} (x) = \bf \dfrac{x - 2}{ - 2} }$

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$\mathbb \color{orange} \underbrace{PENYELESAIAN}$

SOAL 1 :

$\underline{ \overline{ \boxed{ \bold{diketahui}}}}$

f(x) = 2x + 4

$\\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}$

$\sf {f}^{ - 1} (x)$

$\sf {f}^{ - 1} (2)$

$\\ \underline{ \overline{ \boxed{ \bold{jawab}}}}$

=> menentukan f-¹(x) :

$\small \begin{aligned} \tt y &= \tt 2x + 4 \\ \tt 2x &= \tt y - 4 \\ \tt x &= \tt \frac{y - 4}{2} \end{aligned}$

$\small \tt {f}^{ - 1} (x) = \bf \dfrac{x - 4}{2} \\$

=> menentukan f-¹(2) :

$\small \begin{aligned} \tt {f}^{ - 1} (2) &= \tt \frac{2 - 4}{2} \\ &= \tt \frac{ - 2}{2} \\ &= \bf - 1 \end{aligned} \\$

$____________________________________$

$\\$

SOAL 2 :

$\underline{ \overline{ \boxed{ \bold{diketahui}}}}$

$\sf \frac{4x - 7}{7x + 3}$

$\\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}$

$\sf {f}^{ - 1} (x)$

$\\ \underline{ \overline{ \boxed{ \bold{jawab}}}}$

=> menentukan f-¹(x) :

$\small \begin{aligned} \tt y &= \tt \frac{4x - 7}{7x + 3} \\ \tt y(7x + 3) &= \tt 4x - 7 \\ \tt 7xy + 3y &= \tt 4x - 7 \\ \tt 7xy - 4x &= \tt - 3y - 7 \\ \tt x(7y - 4) &= \tt - 3y - 7 \\ \tt x &= \tt \frac{ - 3y - 7}{7y - 4} \end{aligned}$

$\small \tt {f}^{ - 1} (x) = \bf \dfrac{ - 3x - 7}{7x - 4} \\$

$____________________________________$

$\\$

SOAL 3 :

$\underline{ \overline{ \boxed{ \bold{diketahui}}}}$

f(x) = x² - 6x + 15

$\\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}$

$\sf {f}^{ - 1} (x)$
$\sf {f}^{ - 1} (1)$

$\\ \underline{ \overline{ \boxed{ \bold{jawab}}}}$

=> menentukan f-¹(x) :

$\small \begin{aligned} \tt y &= \tt {x}^{2} - 6x + 15 \\ \tt {x}^{2} - 6x &= \tt y - 15 \\ \tt {(x - 3)}^{2} - 9 &= \tt y - 15 \\ \tt {(x - 3)}^{2} &= \tt y - 15 + 9 \\ \tt {(x - 3)}^{2} &= \tt y - 6 \\ \tt x - 3 &= \tt \pm \sqrt{y - 6} \\ \tt x &= \tt 3 \pm \sqrt{y - 6} \end{aligned}$

$\small \tt {f}^{ - 1} (x) = \bf 3 \pm \sqrt{x - 6} \\$

=> menentukan f-¹(1) :

$\small \begin{aligned} \tt {f}^{ - 1} (1) &= \tt 3 + \sqrt{1 - 6} \\ &= \tt 3 + \sqrt{ - 5} \end{aligned} \\$

$\small \begin{aligned} \tt {f}^{ - 1} (1) &= \tt 3 - \sqrt{1 - 6} \\ &= \tt 3 - \sqrt{ - 5} \end{aligned} \\$

$____________________________________$

$\\$

SOAL 4 :

$\underline{ \overline{ \boxed{ \bold{diketahui}}}}$

$\sf g(x) = \frac{2x - 1}{6}$

$\\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}$

$\sf {g}^{ - 1}(x)$

$\\ \underline{ \overline{ \boxed{ \bold{jawab}}}}$

=> menentukan g-¹(x) :

$\small \begin{aligned} \tt y &= \tt \frac{2x - 1}{6} \\ \tt y(6) &= \tt 2x - 1 \\ \tt 6y &= \tt 2x - 1 \\ \tt - 2x &= \tt - 6y - 1 \\ \tt x &= \tt \frac{ - 6y - 1}{ - 2} \end{aligned}$

$\small \tt {g}^{ - 1} (x) = \bf \frac{ - 6x - 1}{ - 2} \\$

$____________________________________$

$\\$

SOAL 5 :

$\underline{ \overline{ \boxed{ \bold{diketahui}}}}$

$\sf f(x) = x - 3$

$\\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}$

$\sf {f}^{ - 1} (x)$

$\\ \underline{ \overline{ \boxed{ \bold{jawab}}}}$

=> menentukan f-¹(x) :

$\small \begin{aligned} \tt y &= \tt x - 3 \\ \tt x &= \tt y + 3 \end{aligned}$

$\small \tt {f}^{ - 1} (x) = \bf x + 3 \\$

$____________________________________$

$\\$

SOAL 6 :

$\underline{ \overline{ \boxed{ \bold{diketahui}}}}$

f(x) = 2 - 2x

$\\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}$

$\sf {f}^{ - 1} (x)$

$\\ \underline{ \overline{ \boxed{ \bold{jawab}}}}$

=> menentukan f-¹(x) :

$\small \begin{aligned} \tt y &= \tt 2 - 2x \\ \tt - 2x &= \tt y - 2 \\ \tt x &= \tt \frac{y - 2}{ - 2} \end{aligned}$

$\small \tt {f}^{ - 1} (x) = \bf \dfrac{x - 2}{ - 2}$

4 votes Thanks 5

feisa21 terima kasih kak

Gusti2601 iya kak, sama² :)

SOAL 1 :

SOAL 2 :

SOAL 3 :

SOAL 4 :

SOAL 5 :

SOAL 6 :

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SOAL 1 :

SOAL 2 :

SOAL 3 :

SOAL 4 :

SOAL 5 :

SOAL 6 :

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