What is the inverse of f(x)= 1/4 x^2+7/4 x-31/16 ?

The inverse of f(x)= 1/4 x^2+7/4 x-31/16 is (-7+4sqrt(x+5))/2 ,-(4sqrt(x+5)+7)/2

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inverse of f(x)= 1/4 x^2+7/4 x-31/16

inverse

f (x) = \frac{1}{4} x^{2} + \frac{7}{4} x - \frac{31}{16}

\frac{- 7 + 4 x + 5}{2}, - \frac{4 x + 5 + 7}{2}

Solution steps

y = \frac{1}{4} x^{2} + \frac{7}{4} x - \frac{31}{16}

Replace

x

with

y

x = \frac{1}{4} y^{2} + \frac{7}{4} y - \frac{31}{16}

Solve for

y, x = \frac{1}{4} y^{2} + \frac{7}{4} y - \frac{31}{16}

\frac{- 7 + 4 x + 5}{2}, - \frac{4 x + 5 + 7}{2}

g (x) = x^{3} + 1

f (x) = \frac{9 x}{2 x - 3}

f (x) = \frac{( x + 2 )}{3 x - 4}

f (x) = \frac{6}{1 + x ^{2}}

f (x) = \frac{2 x}{2 x + 3}

What is the inverse of f(x)= 1/4 x^2+7/4 x-31/16 ?
The inverse of f(x)= 1/4 x^2+7/4 x-31/16 is (-7+4sqrt(x+5))/2 ,-(4sqrt(x+5)+7)/2