What is the domain of f(x)= 1/(\sqrt[3]{7-2x)} ?

The domain of f(x)= 1/(\sqrt[3]{7-2x)} is x 7/2

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domain of f(x)= 1/(\sqrt[3]{7-2x)}

domain

x < \frac{7}{2} or x > \frac{7}{2}

Interval Notation

(- \infty, \frac{7}{2}) \cup (\frac{7}{2}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{7}{2}

The function domain

x < \frac{7}{2} or x > \frac{7}{2}

f (x) = \frac{x + 4}{x - 10}

- 15 x^{2} + 350 x - 2000

f (x) = 1 - sec (2 x) \cdot csc (2 x)

f (x) = 2 - lo g_{5} (x^{2})

\frac{x ^{2} + x + 1}{x ^{2} + 2 x + 1}

What is the domain of f(x)= 1/(\sqrt[3]{7-2x)} ?
The domain of f(x)= 1/(\sqrt[3]{7-2x)} is x< 7/2 \lor x> 7/2