Part a.
The domain is the set of x values such that [tex]x \ge -\frac{1}{2}[/tex], basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve [tex]2x+1 \ge 0[/tex] 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 [tex]\Big[ -\frac{1}{2} , \infty \Big)[/tex] 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
