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

The inverse of f(x)=-1/7 sqrt(16-x^2) is sqrt(-49x^2+16),-sqrt(-49x^2+16)

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inverse of f(x)=-1/7 sqrt(16-x^2)

inverse

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

- 49 x^{2} + 16, - - 49 x^{2} + 16

Solution steps

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

Replace

x

with

y

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

Solve for

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

- 49 x^{2} + 16, - - 49 x^{2} + 16

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

f (x) = 2 ln (x + a)

y = 45 - 0.45 x

f (x) = \frac{t ^{2} + 3 t - 4}{t ^{2} - 6 t + 5}

What is the inverse of f(x)=-1/7 sqrt(16-x^2) ?
The inverse of f(x)=-1/7 sqrt(16-x^2) is sqrt(-49x^2+16),-sqrt(-49x^2+16)