What is the inverse of f(x)= 39/10 t^2 ?

The inverse of f(x)= 39/10 t^2 is sqrt((10t)/(39)),-sqrt((10t)/(39))

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inverse of f(x)= 39/10 t^2

inverse

f (x) = \frac{39}{10} t^{2}

\frac{10 t}{39}, - \frac{10 t}{39}

Solution steps

x = \frac{39}{10} t^{2}

Replace

t

with

x

t = \frac{39}{10} x^{2}

Solve for

x, t = \frac{39}{10} x^{2}

\frac{10 t}{39}, - \frac{10 t}{39}

y = - 9 x - 6

\frac{x - 1}{x ^{2} + 3 x + 2}

f (x) = \frac{e ^{x}}{1 + 7 e ^{x}}

f (x) = 3 (x - 1)^{2} - 5

f (x) = x^{2} - 4 x + 17

What is the inverse of f(x)= 39/10 t^2 ?
The inverse of f(x)= 39/10 t^2 is sqrt((10t)/(39)),-sqrt((10t)/(39))