Question

matematika minat kelas 11​

Answer 1

Diketahui

[tex] \sf \: f(2x - 5) = x \sqrt{x - 1} [/tex]

[tex] \sf \: f \: turunan \: pertama \: [/tex]

Ditanya

[tex] \sf12f(15) = ....[/tex]

Jawab

[tex] \sf \: f(2x - 5) = x \sqrt{x - 1} [/tex]

[tex] \sf \: f(x) = ( \frac{x + 5}{2} ) \sqrt{ \frac{x + 5}{2} - 1 } [/tex]

[tex] \sf \: f(x) = ( \frac{x + 5}{2} ) \sqrt{ \frac{x + 5 - 1}{2} } [/tex]

[tex] \sf \: \: f(x) = ( \frac{x + 5}{2} ) \sqrt{ \frac{x + 3}{2} } [/tex]

[tex] \sf \: f’(x) = \frac{1}{2} \sqrt{ \frac{x + 3}{2} } + \frac{1}{ 2\sqrt{ \frac{x + 3}{2} } } \times \frac{1}{2} ( \frac{x + 5}{2} )[/tex]

[tex] \sf \: f’(15) = \frac{14}{6} [/tex]

[tex] \sf \:12 f’(15) = 12\frac{14}{6} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = 28[/tex]

Answer 2

Jawaban:

2f-¹(15) = 360

Penjelasan dengan langkah-langkah:

[tex]f(2x - 5) = x \sqrt{x - 1} \\ 2x - 5 = y \\ 2x = y + 5 \\ x = \frac{y + 5}{2} \\ {f}^{ - 1} (y) = \frac{y + 5}{2} \sqrt{ \frac{y + 5}{2} - 1} \\ {f}^{ - 1} (x) = \frac{x + 5}{2} \sqrt{ \frac{x + 5}{2} - 1 } \\ {f}^{ - 1} (15) = \frac{15 + 5}{2} \sqrt{ \frac{15 + 5}{2} - 1} \\ = 10 \sqrt{10 - 1} \\ = 10 \sqrt{9} \\ = 30 \\ 12 {f}^{ - 1} (15) = 12 \times 30 \\ = 360[/tex]