For f(x) = (2x+5)/[√(x2) − 1]:
f(5) = (2(5)+5)/[√(52) − 1] or 15/[5 − 1] or 15/4 or 3 + 3/(5 − 1).
f(a) = (2a+5)/[√(a2) − 1] or (2a+5)/[a − 1] or 2 + 7/[a − 1].
f(x+h) = (2(x+h)+5)/[√{(x+h)2} − 1] or (2x+2h+5)/(x+h−1) or 2 + 7/(x+h−1).
Jake P.
asked 01/22/19Consider the function f(x) = 2x + 5 / √ x^2 − 1 . Find f(5), f(a), f(x + h).
For f(x) = (2x+5)/[√(x2) − 1]:
f(5) = (2(5)+5)/[√(52) − 1] or 15/[5 − 1] or 15/4 or 3 + 3/(5 − 1).
f(a) = (2a+5)/[√(a2) − 1] or (2a+5)/[a − 1] or 2 + 7/[a − 1].
f(x+h) = (2(x+h)+5)/[√{(x+h)2} − 1] or (2x+2h+5)/(x+h−1) or 2 + 7/(x+h−1).
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