1. (x,f(x)) and (3,f(3)) 2. (x,f(x)) and (1,f(1)) 3. (x+1,f(x+1)) and (x-1,f(x-1)) 4. (x,f(x)) and (x+h,f(x+h)) 5. (x,f(x)) and (x+2,f(x+2))For f(x)=x², find the slope of the line containing...

1. (x,f(x)) and (3,f(3))

2. (x,f(x)) and (1,f(1))

3. (x+1,f(x+1)) and (x-1,f(x-1))

4. (x,f(x)) and (x+h,f(x+h))

5. (x,f(x)) and (x+2,f(x+2))

For f(x)=x², find the slope of the line containing the preceding points.

Also, I need help finding the following limits.

Lim as:

 x→3 (x²-9)/(x-3)

x→1 (x²-1)/(x-1)

x→-1 (x+1)²-(x-1)²/2

I need to make a conjecture about the slope of the secant line connecting two very, very, very close points on x², then use the generalization to answer the following.

h→0 [(2+a)²-2²]/a


Expert Answers
magnesi eNotes educator| Certified Educator

Answer to your Limits Question:

1) If you Factor the Numerator you will get (x-3)(x+3)/(x-3) allowing you to cancel the (x-3).  You will then be left with (x+3). Using substitution you can conclude that (3+3) = 6. The limit as x goes to 3 of (x^2-9)/(x-3) is 6

2) Using the same method as above, factoring, you will get (x-1)(x+1)/(x-1) and can cancel the (x-1) leaving you with (x+1). Substituting 1 for x will result in 2. The Limit as x approaches 1 of (x^2-1)/(x-1)=2

3) You must FOIL the numerator of this problem leaving you with (4x)/(2).  Again using substitution (placing -1 in for x) you get (-4)/2=-2. The Limit as x approaches -1 of (x+1)^2-(x-1)^2/2=-2

neela | Student


slope = (f(x)-f(3)/(x-3) = (x^2- 3^2/(x-3) = (x+3)(x-3)/(x-3)= x+3


slope is (x^2-1^2)/(x-1) =(x+1)(x-1) = x+1


[(x+1)^2-(x-1)^2]/[(x+1-(x-1)]/2 =(2x)(2)/ = 4x/2 = 2x


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

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


[(x^2-(x+2)^2]/(0-2) = -(2*2x+2^2)/2 = 2x+2




x→3 (x²-9)/(x-3)

x-->1 (x+3)(x-3)/(x-3) = x-->1 Lt(x+3) = 1+3=4

2)x→1  (x²-1)/(x-1)

x-->lt(x+1)(x-1)/x-1) = x-->1 lt(x+1) =1+1=2


x→-1 (x+1)²-(x-1)²/2

=x-->-1 lt{x^2+2x+1-(x^2-2x+1)}/2 = x-->-1 lt(4x)/2 =4(-1)/2 = -2.

sec A = sqrt[1+(tanA)^2 ]= sqrt{1+[(d/dx) of x^2]} = sqrt{1+(2x)^2) =sqrt (1+4x^2) at x.

a-->0 Lt [(2+a)-2^2]

= a->0 [2^2+2a+a^2)- 2^2]/a

= a--> LT(2+a) = 2