What is the domain of f(x)=sqrt(16-t^2) ?

The domain of f(x)=sqrt(16-t^2) is -4<= t<= 4

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domain of f(x)=sqrt(16-t^2)

domain

f (x) = 16 - t^{2}

- 4 \leq t \leq 4

Interval Notation

[- 4, 4]

Solution steps

Find non-negative values for radicals:

- 4 \leq t \leq 4

The function domain

- 4 \leq t \leq 4

y = 2 e^{x - 2}

y - 5 = - 3 (x - 1)

f (x) = ln (1 + x^{3})

y = x + 5

f (x) = π - x

What is the domain of f(x)=sqrt(16-t^2) ?
The domain of f(x)=sqrt(16-t^2) is -4<= t<= 4