سلام
دو به توان ۹ از کجا اومد
و یکبار روش حل توضیح میدید
نمایش کامل متن تصویر (کلیک کنید)
$f(1/2) = 1 \implies 1 = \sqrt[3]{2/2 + b} \implies 2/2 + b = 1 \implies a/2 + b = 0 \implies b = -a/2 (*)$
$f^{-1}(8) = 5 \implies f(5) = 8 \implies 8 = \sqrt[3]{5a + b}$
$5a + b = 2^3 \implies 5a + b = 8 \xrightarrow{(*)} 5a - a/2 = 8 \implies 9a/2 = 8 \implies a = 2 \xrightarrow{(*)} b = -1$
$a - b = 2 - (-1) = 3$
![$f(1/2) = 1 \implies 1 = \sqrt[3]{2/2 + b} \implies 2/2 + b = 1 \implies a/2 + b = 0 \implies b = -a/2 (*)$
$f^{-1}(8) = 5 \implies f(5) = 8 \implies 8 = \sqrt[3]{5a + b}$
$5a + b = 2^3 \implies 5a + b = 8 \xrightarrow{(*)} 5a - a/2 = 8 \implies 9a/2 = 8 \implies a = 2 \xrightarrow{(*)} b = -1$
$a - b = 2 - (-1) = 3$](https://services.porsan.app/files//Question/342885.jpg)
