+0

# Function

0
57
2

Compute the domain of the real-valued function f(x) = sqrt(3 + sqrt(5 + sqrt(x)).

Oct 10, 2022

#1
+1009
+2

$$f(x)=\sqrt{3 + \sqrt{5 + \sqrt{x}}}$$

$$x\ge0$$

Smallest value of the function would be $$\sqrt{3 + \sqrt{5}}$$

Maximum value is infinity.

$$a + b\sqrt{c} = \sqrt{3 + \sqrt{5}}$$

$$a + b\sqrt{5} = \sqrt{3 + \sqrt{5}}$$

(a + bsqrt(5))^2 = 3 + sqrt(5)

a^2 + 5b^2 + 2absqrt(5) = 3 + sqrt(5)

a^2 + 5b^2 = 3

sqrt(5) = 2absqrt(5)

2ab = 1

ab = 1/2

You can do it from here... Domain: [a + bsqrt(c), inf)

Get a and b, or just use the sqrt(3 + sqrt(5))

Oct 10, 2022