Question

For f(x)=√(2x+1) , find the following:

a. The domain of f(x)

b. f(10)

c. f(x+2a)

Explain

Answer 1

Part a.

The domain is the set of x values such that
`x \ge -(1)/(2)`, basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve
`2x+1 \ge 0` for x (subtract 1 from both sides; then divide both sides by 2). I set 2x+1 larger or equal to 0 because we want to avoid the stuff under the square root to be negative.

If you want the domain in interval notation, then it would be
`\Big[ -(1)/(2) , \infty \Big)` which means the interval starts at -1/2 (including -1/2) and then it stops at infinity. So technically it never stops and goes on forever to the right.

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Part b.

I'm going to use "sqrt" as shorthand for "square root"

f(x) = sqrt(2x+1)

f(10) = sqrt(2*10+1) ... every x replaced by 10

f(10) = sqrt(20+1)

f(10) = sqrt(21)

f(10) = 4.58257569 which is approximate

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Part c.

f(x) = sqrt(2x+1)

f(x) = sqrt(2(x)+1)

f(x+2a) = sqrt(2(x+2a)+1) ... every x replaced by (x+2a)

f(x+2a) = sqrt(2x+4a+1) .... distribute

we can't simplify any further