Tangent Line Curves And Limit, Looking To Check My Answers?

Question

1.Find The Points On The Curve y = x^3 - x^2 - x + 1 where the tangent line is horizontal.

2. How many tangent lines to the curve y = x/(x+1) pass through the point (1,2)? At which points do these tangent lines touch the curve?

3. Evaluate lim x--> 1 x^ 9 - 1 / x - 1

lim x^9 - 1 / x - 1
x--> 1

Answer 1

L'Hopital's rule
lim x^ 9 - 1 / x - 1
= lim 9x^8 / 1
= 9x^8 as x = 1
So limit = 9

For Q2:
y = x / (x + 1) = 1 - 1/(x+1)
dy/dx = 1 / (x+1)^2
Consider a general point on the curve (p, q)
dy/dx at that point = 1 / (p+1)^2
q = 1 - 1/(p+1)

Equation of tangent is:
y - q = m*(x - p)
y - 1 + 1/(p+1) = 1 / (p+1)^2 * (x - p)
Does this pass through the point (1,2). Let's set y = 2, x = 1
2 - 1 + 1/(p+1) = 1 / (p+1)^2 * (1 - p)
1 + 1/(p+1) = (1 - p) / (p+1)^2
(p + 2) / (p+1) = (1 - p) / (p+1)^2
(p+2)*(p+1) = 1 - p
p^2 + 3p + 2 = 1 - p
p^2 + 4p + 1 = 0

The solutions to this equation are:
p = -2 +/- sqrt(3)

So the answer is 2 points. Their x cords are the p values above. Their y cords will bethe corresponding q values. That's an exercise for you to find.

Answer 2

for part 2 there are 2 tangent lines to the graph that pass thru (1,2)

step 1: find the slope of the tan lines to any point on the graph by taking first deritave so:

m=1/(x+1)²

step 2....now use the point slope formula for a line y-y1=m(x-x1)
y1=2 x1=1 m from eqn above
y-2=(1/(x+1)²)(x-1)

we know the other point must lie on the curve y=x/(x+1) so that eqn can be subed in for y:

(x/(x+1))-2=(1/(x+1)²)(x-1)

solving for x gives x= -√3-2 and x=√3-2

now sub these x-values into the eqn for the curve to find the points where the tangent lines touch the curve. To check I did graph these on the computer and the x-values are correct.

Answer 3

y´=3x^2-2x-1=0
x=((2+-sqrt(4+12))/6
so x= 1 and x= -1/3 (get y by substitution)

2)y´= 1/(x+1)^2
The tangent at a point (a,a/(a+1)) is
y-a/(a+1)=1/(a+1)^2 (x-a)Passing through (1,2)
2-a/(a+1)=1/(a+1)^2 (1-a)
a+2= 1/(a+1) *(1-a) and
a^2+3a+2=1-a
a^2+4a+1=0 so a= ((-4+-sqrt(12))/2
=-2+-sqrt3 (two points) (get y from the function
3)Factoring x^9-1= (x-1)(x^8+x^7+x^6+++1)
so the limit =9

Answer 4

1. Differentiate the equation to get y'=3x^2-2x-1
set this equal to zero (for turning points)
3x^2-2x-1 = 0
By using quadratic formula or factorising
Hence x= 1 or -1/3
Hence y = 0 or 1.185 (with all decimal places recurring)


(1, 0) and (-1/3, 1.185...)
2. ?

3. 9

Answer 5

d/dx e^x = e^x The tangent has slope e^x, passes with the aid of (x, e^x) y = m x + c e^x = x e^x + c (c is y intercept) c = e^x (a million - x) resolve c = 0 ? x = a million So tangent to curl that passes with the aid of beginning is y = e x verify: y(0) = 0, passes with the aid of beginning. y(a million) = e = e^a million

Answer 6

1)
dy/dx must = 0
3x^2 - 2x-1 =0
x= -1/3, f(-1/3) = 32/27
x=1, f(1)= 0
(-1/3,32/27) and (1,0)