yes that's what I meant, $$\frac{cos(h)-1}{h},$$
I know that it is in fact the derivative of sinx but when I input the values it tells me in the calculator I'm not getting the same.
I think it's cos(h-1) the question has it written the way I wrote it. H is the same as ΔX because it is a limit. Okay, but is my assumption that the value should be equivalent to sin correct?
Homework Statement
Use a calculator to evaluate the quantity (cosh-1) / h for h = 0.1, 0.01, 0.001, and 0.0001.
Estimate the value of the limit of h -> 0 of (cos h -1) / h
Homework Equations
The Attempt at a Solution
okay so at 0.1 it is 1.523 x 10^-5 and it goes on like that...
Unredeemed, can you please show me how to go on with the process you are talking about? We were taught to find the distance formula then differentiate and solve when it is 0. I've never studies Lagrange's multipliers.
I don't understand what you mean. We were just taught to use the distance formula D^2 = (X2-X1)^2 + (Y2-Y1)2
I subbed in -2 as x1 and 4 as y1 then the equation of x= y^2 -8y + 18 as X2. Am I doing it wrong?
Homework Statement
Find the point(s) on the parabola x= y^2-8y+18 closest to (-2,4)
Homework Equations
The Attempt at a Solution
Using the distance formula I found that d^2 = y^4 - 16y^3 + 105y^2 - 328y + 416
after differentiating I have 2d' = 4y^3 - 48y^2 + 210y -328
I'm not...
Okay so I simplified that farther to achieve this, 2(3x-5) / x-2 > 0. So now how do I i know if X> 5/3 OR X<5/3, also wouldn't there be an inequality in the denominator, such as x>2 or x<2 . Also how do display the mathematical operation like that on the forum?
Homework Statement
lx/(x-2)l < 5
Homework Equations
The Attempt at a Solution
x/(x-2) < 5
x< 5x-10
10 < 4x
5/2 < x
x/(x-2) > -5
x > -5x+10
6x > 10
x > 5/3
The answer is x < 5/3 and x > 5/2
so where did I go wrong on the second one?
I got a ti84, and i can draw a tangent line and it'll tell me the formula, so when I drew a tangent line through -5, the line was not horizontal completely. If you plug in -5 into 2x^2 + 10x= 0, it works out, but when you plug it back into the first one you don't get 0.
Homework Statement
At what points on the curve y = (x^2)/(2x+5) is the tangent line horizontal?
Homework Equations
Quotient rule
The Attempt at a Solution
I figured out the derivative which is
2x(2x+5) - 2x^2
-----------------
(2x+5)^2
I also know that for the equation of...
Well we never learned l'hopital's rule yet but after googling it and attempting it i got
(1/2) - 1
----------
(3/2) - (1/2)
which gave me -1/4, not the answer.
Now as for jbunnii's plan. I got up to
- x ( (√3x+1) + (√2x+4) )
------------------------
(x-3) ((√x+1) + (√2x+1))
I'm unsure...
Homework Statement
Evaluate lim x →0 √(x+1) - √(2x+1)
-----------------
√(3x+4) - √(2x+4)
Homework Equations
The Attempt at a Solution
First I rationalized the numerator by multiplying everything by √(x+1) + √(2x+1) / √(x+1)...