What is the inverse of f(x)=((x^5)/7)^{1/7}-1 ?

The inverse of f(x)=((x^5)/7)^{1/7}-1 is \sqrt[5]{7x^7+49x^6+147x^5+245x^4+245x^3+147x^2+49x+7}

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inverse of f(x)=((x^5)/7)^{1/7}-1

inverse

f (x) = (\frac{x ^{5}}{7})^{\frac{1}{7}} - 1

Solution steps

y = (\frac{x ^{5}}{7})^{\frac{1}{7}} - 1

Replace

x

with

y

x = (\frac{y ^{5}}{7})^{\frac{1}{7}} - 1

Solve for

y, x = (\frac{y ^{5}}{7})^{\frac{1}{7}} - 1

f (x) = 5 (x + 4)

e^{x - 1} + 2

f (x) = \frac{3 x + 5}{4}

sin (z)

f (x) = - (x - 2)^{2}

What is the inverse of f(x)=((x^5)/7)^{1/7}-1 ?
The inverse of f(x)=((x^5)/7)^{1/7}-1 is \sqrt[5]{7x^7+49x^6+147x^5+245x^4+245x^3+147x^2+49x+7}