What is the inverse of f(x)= 1/7 x^2+3 ?

The inverse of f(x)= 1/7 x^2+3 is sqrt(7x-21),-sqrt(7x-21)

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inverse of f(x)= 1/7 x^2+3

inverse

f (x) = \frac{1}{7} x^{2} + 3

7 x - 21, - 7 x - 21

Solution steps

y = \frac{1}{7} x^{2} + 3

Replace

x

with

y

x = \frac{1}{7} y^{2} + 3

Solve for

y, x = \frac{1}{7} y^{2} + 3

7 x - 21, - 7 x - 21

f (x) = x^{2 - 10 + 15}

+ \frac{z ^{- 1}}{( 1 - 0.5 \cdot z ^{- 1} ) ^{2}}

y = \frac{1}{2 x - 4}

f (x) = \frac{x ^{2} + 3 x - 10}{x - 2}

e^{3 y}

What is the inverse of f(x)= 1/7 x^2+3 ?
The inverse of f(x)= 1/7 x^2+3 is sqrt(7x-21),-sqrt(7x-21)