What is the inverse of \sqrt[4]{-8-8x},x<=-1 ?

The inverse of \sqrt[4]{-8-8x},x<=-1 is -(x^4+8)/8

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inverse of \sqrt[4]{-8-8x},x<=-1

inverse

4 - 8 - 8 x, x \leq - 1

- \frac{x ^{4} + 8}{8}

Solution steps

y = 4 - 8 - 8 x

Replace

x

with

y

x = 4 - 8 - 8 y

Solve for

y, x = 4 - 8 - 8 y

- \frac{x ^{4} + 8}{8}

f (x) = \frac{8 x - 7}{x + 6}

\frac{( x + 3 )}{x + 10}

- 3 tan (3 π x)

f (x) = 5 x^{2} \frac{1}{2}

13 - x

What is the inverse of \sqrt[4]{-8-8x},x<=-1 ?
The inverse of \sqrt[4]{-8-8x},x<=-1 is -(x^4+8)/8