Question

Find the domain of the real valued function: f(x) = $\frac{1}{\sqrt[3]{x - 2} . log_{(4-x)} 10}$

Answer 1

Answer:

Domain of f(x): x < 2 or 2 < x < 3 or 3 < x < 4

interval notation:(-∞,2) ∪ (2,3) ∪ (3,4)

Step-by-step explanation:

To find the domain of the real valued function:

$f(x) =\frac{1}{\sqrt[3]{x - 2}.log_{(4-x)} 10}$

as domain is a set of input values for which is real and defined.

first we had to check values for logarithmic base: x < 3 or 3 < x < 4

$log_{f(x)} a => 0 < f(x) < 1\:\:or\:\:f(x) > 1$

0 < (4-x) < 1  or (4-x) > 1; x < 3 or 3 < x < 4

so for 4-x > 1; x < 3

x < 3 or 3 < x < 4

for :∛(x-2) log₄-ₓ (10) =0; on putting x=2

so x=2 is undefined

so

Domain of f(x): x < 2 or 2 < x < 3 or 3 < x < 4

interval notation:(-∞,2) ∪ (2,3) ∪ (3,4)