Calculus
f−1(x)=17x−57f^{-1}\(\left\)(x\(\right\))=\(\frac\)17x-\(\frac\)57
f−1(x)=17x+57f^{-1}\(\left\)(x\(\right\))=\(\frac\)17x+\(\frac\)57
f−1(x)=17x+5f^{-1}\(\left\)(x\(\right\))=\(\frac\)17x+5
f−1(x)=17x−5f^{-1}\(\left\)(x\(\right\))=\(\frac\)17x-5