Question

1. Consider the curve y=x^2.

a. write down (dy)/(dx)
My answer: 2x

The point P(3,9) lies on the curve y=x^2.

b. Find the gradient of the tangent to the curve at P.
My answer: 2*3=6

c. Find the equation of the normal to the curve at P. Give your answer in the form y=mx+c.
My answer: y=6x-9 but the answer key says y=-1/6x+9.5. Why and how??

2. f(x)=5x^3-3x^5+1 for -1.5<=x<=1.5 and -6<=y<=6.

a. Write down f'(x)
My answer: f'(x)=15x^2-15x^4

b. Find the equation of the tangent to the graph of y=f(x) at (1,3).

My answer:y=x+1 but the answer key says f'(1)=0
Why and how? Isn't asking for EQUATION?

c. Write down the coordinates of the second point where this tangent intersects the graph of y=f(x).
I don't understand what this is asking and am stuck!




3. A small manufacturing company makes and sells x machines each month. The monthly cost C, in dollars, of making x machines is given by
C(x)= 2600+0.4x^2

The monthly income I, in dollars, obtained by selling x machines is given by
I(x)=150x-0.6x^2

P(x) is the monthly profit obtained by selling x machines.

a. Find P(x).
b. Find the number of machines that should be made and sold each month to maximize P(x).
c. Use yours answer to part b to find the selling price of each machine in order to maximize P(x).

Answer 1

1. a) and b) are correct

c) the normal is perpendicular to the tangent
So if the slope of the tangent at (3,9) is 6
then the slope of the normal is -1/6
y = (-1/6)x + b, plug in the point (3,9)
9 = (-1/6)(3) + b
b = 19/2

y = (-1/6)x + 19/2 or y = (-1/6)x + 9.5

2. a) good!
g) if f ' (x) = 15x^2 - 15x^4 , then
f '(1) = 15-15 = 0
so slope = 0, which is a horizontal line
(You have a slope of 1)

A horizontal line through the point is (1,3)
is y = 3 or f(x) = 3

I don't see your answers to #3

Answer 2

for #2c, there is a second point at about (-1.38,3)

see the graph at

http://www.wolframalpha.com/input/?i=solve+5x^3-3x^5%2B1+%3D+3