What is the domain of y=12+log_{10}(1-x^2) ?

The domain of y=12+log_{10}(1-x^2) is -1<x<1

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domain of y=12+log_{10}(1-x^2)

domain

y = 12 + lo g_{10} (1 - x^{2})

- 1 < x < 1

Interval Notation

(- 1, 1)

Solution steps

Find positive values for logs:

- 1 < x < 1

The function domain

- 1 < x < 1

f (x) = \frac{( - 3 - x )}{5 x - 8}

\frac{x - 3}{5}

f (x) = 1 1^{x + 3} - 7

f (x) = x^{2} + y^{2} - 2 x - 3 y - 6 = 0

\frac{6 x ^{2} + x}{x ^{2} + 6}

What is the domain of y=12+log_{10}(1-x^2) ?
The domain of y=12+log_{10}(1-x^2) is -1<x<1