What is the inverse of f(x)=x^3+4x^3+1 ?

The inverse of f(x)=x^3+4x^3+1 is cube root of (x-1)/5

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inverse of f(x)=x^3+4x^3+1

inverse

f (x) = x^{3} + 4 x^{3} + 1

3 \frac{x - 1}{5}

Solution steps

y = x^{3} + 4 x^{3} + 1

Replace

x

with

y

x = y^{3} + 4 y^{3} + 1

Solve for

y, x = y^{3} + 4 y^{3} + 1

3 \frac{x - 1}{5}

f (x) = n (n + 1)

f (x) = - 6 (x - 3)^{2} - 2

f (x) = \frac{3 x - 2}{x - 5}

f (x) = 4 \cdot lo g_{\frac{1}{2}} (- x + 10)

f (x) = \frac{x ^{2} - 1}{x ^{2} - 9}

What is the inverse of f(x)=x^3+4x^3+1 ?
The inverse of f(x)=x^3+4x^3+1 is cube root of (x-1)/5