What is the inverse of 1/4 (t^4+8) ?

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

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inverse of 1/4 (t^4+8)

inverse

\frac{1}{4} (t^{4} + 8)

4 4 t - 8, - 4 4 t - 8

Solution steps

y = \frac{1}{4} (t^{4} + 8)

Replace

t

with

y

t = \frac{1}{4} (y^{4} + 8)

Solve for

y, t = \frac{1}{4} (y^{4} + 8)

4 4 t - 8, - 4 4 t - 8

f (x) = \frac{3}{2} cos (\frac{x}{3}) - \frac{1}{2}

f (x) = a 3 - x

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

y = \frac{1}{x - 2} - 3

f (x) = \frac{( 6 x + 2 )}{x}

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