What is the domain of f(t)=sqrt(2t^2+5t-3) ?

The domain of f(t)=sqrt(2t^2+5t-3) is t = 1/2

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domain of f(t)=sqrt(2t^2+5t-3)

domain

f (t) = 2 t^{2} + 5 t - 3

t \leq - 3 or t \geq \frac{1}{2}

Interval Notation

(- \infty, - 3] \cup [\frac{1}{2}, \infty)

Solution steps

Find non-negative values for radicals:

t \leq - 3 or t \geq \frac{1}{2}

The function domain

t \leq - 3 or t \geq \frac{1}{2}

- \frac{4 x}{( x - 7 ) ( x + 5 )}

f (x) = - 1 x^{2} + 4

y, y \leq x + 4

62.5 x + 460

f (x) = x^{4} - 2 x^{3} + x^{2}

What is the domain of f(t)=sqrt(2t^2+5t-3) ?
The domain of f(t)=sqrt(2t^2+5t-3) is t<=-3\lor t>= 1/2