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

The domain of f(x)=sqrt(-2x^2+16x) is 0<= x<= 8

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

domain

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

0 \leq x \leq 8

Interval Notation

[0, 8]

Solution steps

Find non-negative values for radicals:

0 \leq x \leq 8

The function domain

0 \leq x \leq 8

f (x) = \frac{x}{∣ x + 4 ∣ - 10}

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

f (x) = 2020 + 2 - x

f (x) = 2 e^{\frac{x ^{2} - 1}{4 x}}

\frac{x + 1}{8 x - 9}

What is the domain of f(x)=sqrt(-2x^2+16x) ?
The domain of f(x)=sqrt(-2x^2+16x) is 0<= x<= 8