Simplify [f(x+h)-f(x)]/h, where (a) f(x) = 2x+3, (b) f(x)=1/(x+1), (c) f(x)=x^2?

Question

I can't figure out how to solve these three, please help. Thanks!

Answer 1

This looks like the formula for finding a derivative using limits. For a:

[2(x+h) + 3 - (2x +3)] / h
=[2x + 2h - 2x + 3 - 3] / h
= 2h/h = 2


For b:
(1/ (x + h +1)) - (1/(x+1)) / h
-> multiply the first term in the numerator by (x+1), the second by (x+h+1).
= [(x+1) - (x+h +1)]/[h(x+h+1)(x+1)]
= -h / h(x + h + 1)(x+1)
= -1 / [(x+h+1)(x+1)]
= -1/ (x^2 + 2x + xh + 1)

For c:
[(x+h)^2 - x^2] / h
= [x^2 + 2xh + h^2 - x^2] / h
= [2xh + h^2] / h
= 2x + h

If you are taking the limit on these as h -> 0
a: 2
b: -1/ (x^2 + 2x + 1) = -1 / (x + 1)^2
c: 2x

Answer 2

As stated, the formula presented is that of the derivative; the statement that is missing is the limit representation:

lim (h goes to zero) [f(x+h)-f(x)]/h

a) f(x) = 2x+3 substitute as follows

f(x+h) = 2(x+h) - 3 = 2x + 2h -3
f(x) = 2x+3

f(x+h) - f(x) = 2h

divided by h = 2h / h = 2

b) f(x+h) = 1/(x+h+1)
f(x) = 1/ (x+1)

f(x+h) - f(x) = 1/(x+h+1) - 1/(x+1)

= [(x+1) - (x+h+1)]/[(x+h+1)(x+1)]
= -h/[(x+h+1)(x+1)]

Divide by h and let h go to zero on the limit

solution = -1 / [(x+1)(x+1)]

c) f(x+h) = (x+h)(x+h) = x^2 +2xh + h^2
f(x) = x^2

f(x+h) - f(x) = 2xh + h^2

divide by h and let it go to zero on the limit

Solution= 2x + h
Solution = 2x

Please review the limit definition and practice, practice, practice

Answer 3

a.)
f(x) = 2x + 3

f(x + h) = 2(x + h) + 3 = 2x + 2h + 3

[f(x + h) - f(x)]/h

[(2x + 2h + 3) - (2x + 3)]/h
[2x + 2h + 3 - 2x - 3]/h
[2h]/h

ANS : 2

-------------------------------------------------------------------

b)
f(x) = 1/(x + 1)

f(x + h) = 1/((x + h) + 1)
f(x + h) = 1/(x + h + 1)

[f(x + h) - f(x)]/h

[(1/(x + h + 1)) - (1/(x + 1))]/h
[(1/((x + h) + 1))) - (1/(x + 1))]/h
[((x + 1) - (x + h + 1))/((x + 1)(x + h + 1))]/h
[(x + 1 - x - h - 1)/((x + 1)(x + h + 1))]/h
[(-h)/((x + 1)(x + h + 1))]/h
[(-h)/((x + 1)(x + h + 1))]/(h/1)
[(-h)/((x + 1)(x + h + 1))]*(1/h)
(-h)/(h(x + 1)(x + h + 1))
-1/((x + 1)(x + h + 1))
-1/(x^2 + xh + x + x + h + 1)
-1/(x^2 + 2x + xh + h + 1)
-1/(x^2 + (2x + xh) + (h + 1))
-1/(x^2 + (2 + h)x + (h + 1))

-----------------------------------------------------

f(x + h) = (x + h)^2
f(x + h) = (x + h)(x + h)
f(x + h) = x^2 + 2xh + h^2

[f(x + h) - f(x)]/h

[(x^2 + 2xh + h^2) - x^2]/h
[x^2 + 2xh + h^2 - x^2]/h
[2xh + h^2]/h
[h(2x + h)]/h

ANS : 2x + h

Answer 4

first, this is the formula for the slope of the secant line between two points on a graph, also called the difference quotient... it's not the derivative, just the first step in getting the derivative directly from its definition

so set up the quotient for each function, and simplify

remember that to get f(x + h) you have to replace each instance of x in the formula for f(x) by x + h... that's the place where most of my students fail to do one of these correctly

for part (b) you will get a "complex fraction" to simplify... of the three functons, that one will make you do the most complicated algebra

Answer 5

f(x)=52x^2+12