2 Questions?

Question

I have a class on monday and I'm freaking out about some of these questions on the assignment.

1)Solve: ln(x^3+x^2-4x-4)-ln(x+1)=5

2)What is the slope of the tangent to the graph of y=x^3-7x^2+14x-8 at the point (3,-2)?


Thanks alot, any help appreciated.

Answer 1

1st Answer is x = (plus or minus) sqrt (e^5 + 4)

"sqrt" means the "square root"
& "plus or minus" means you actually have 2 solutions here
& to understand how to begin the work is to realize that
ln A - ln B = ln (A/B), in other words a subtraction of logs simplifies to a DIVISION...Now look up "division of polynomials" in ur textbook.
Practice working it out...you will get
(x^3 + x^2 - 4x - 4) / (x + 1) = x^2 - 4.

So now you have ln ( x^2 - 4) = 5, so take the "exponential" (that means "e" ) of both sides... you should get
x^2 - 4 = e^5, so now add "4" to both sides... do you get
x^2 = 4 + e^5 ? Now take "square root" of both sides...Ta-Daaa :-)

2nd Answer is "y=-x+1" (Start by taking the DERIVATIVE of
y=x^3 - 7x^2 + 14x - 8... Do you get

y' = 3x^2 - 14x + 14 ? (You should.)
Then plug in x=3, so y'(3) = 27 - 3*14 + 14 = 27-28 = -1...this is your slope (In the line y=mx+b, you've just calculated that the slope, which is "m," so m=-1).

So you put this information into y=mx+b & it becomes y=-1x+b,
or just y= -x+b...Now to get the "b," just use the given "point" information: the point (3,-2) always means =(x,y), so here
x=3 and y=-2... Put it in...you get -2=-3+b, so now you know
b=1 (:-))... So put it in & you get "y=-x+1"

Let me know if there was anything in the above that i need to explain better & Good Luck on Monday !!

Answer 2

1) Solve: ln(x^3+x^2-4x-4) - ln(x+1) = 5

ln(x^3+x^2-4x-4) - ln(x+1) = 5
ln{(x^3+x^2-4x-4)/(x+1)} = 5
ln(x² - 4) = 5

Exponentiating

(x² - 4) = e^5
x² = e^5 + 4
x = ±√(e^5 + 4)

2) What is the slope of the tangent to the graph of
y=x^3-7x^2+14x-8 at the point (3,-2)?

Take the derivative to find the slope.

dy/dx = 3x² - 14x + 14

Plug in the value x = 3

dy/dx = 3*9 - 14*3 + 14 = 27 - 42 + 14 = -1

Answer 3

First one... ln a - ln b = ln (a/b)

So you get:

ln [(x^3 + x^2 - 4x - 4)/(x+1)] = 5

do "e" on both sides to cancel the ln...

(x^3 + x^2 - 4x - 4)/(x+1) = e^5

Now, you can probably factor the fraction to make it nicer...
It factors to (x+1)(x^2 - 4), so the (x+1) on top and bottom cancel out...

You're left with
x^2 - 4 = e^5
x^2 = e^5 + 4
x = + or - sqrt(e^5 + 4)
-------------
Number two... I can't quite remember how to do this... I know you need to take the derivative of the equation, but I don't remember what to do after that. (Haven't use calculus in 12+ years...)

(thanks to the person who posted below me!!)

The derivative would be:
3x^2 - 14x + 14
So...
3(3^2) - 14(3) + 14
= -1

Answer 4

2)What is the slope of the tangent to the graph of
y = x^3 - 7x^2 + 14x - 8 at the point (3,-2)?
y' = 3x^2 - 14x + 14
y'(3) = 3*9 - 14*3 + 14
y'(3) = 27 - 28 = -1

Answer 5

1. Rewrite the ln using ln(z/m)= lnz)-ln(m). Then
see if the expression is divisible by (x+1). If so,
you've got a quadratic, which you know how to solve. If not, just solve the resulting cubic. Beware of multiple solutions.

2. Just take the derivative. Plug in x=3, Evaluate the
derivative. Bob's your uncle.