what do you mean by "solve"?
f(x) = √(2x+1)
doesn't involve a difference quotient.
If you want to write down the difference quotient, then that would be
(f(x+h)-f(x))/h
= (√(2(x+h)+1) - √(2x+1))/h
= (√(2x+1+2h) - √(2x+1))/h
Not sure where you want to go with that . . .
How do you solve f(x)=square root of 2x+1 using the difference quotient? I need to see the steps.
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