CPhill

By a property of absolute values.... l a l * l bl  = l a * b l

So

l z1 l / l z2 l =

l z1 l * l 1 / z2 l =

l z1 * (1 / z2) l =

l (z1 * 1 ) / z2 l =

l z1 / z2 l

And the RHS = LHS

Note that cosxcscx = cotx    so we have

F(x) = [(x² + 1)*cotx] /([3 - cotx] =  [x²+1][cotx][3 - cotx]^-1

And since we have three functions - call them u, v and w - we can write F(x) = u*v*w...so F '(x) = u'*v*w + u*v'*w + u*v*w'

So we have.....using the product and chain rules......

F '(x) = (2x) [x²+1][cotx][3 - cotx]^-1 - [x²+1][cscx]² [3 - cotx]^-1 - [x²+1][cotx][3 - cotx]^-2 [cscx]²

And we could write this in several different ways.......here's one .......

F '(x) = (2x) [x²+1][cosx][cscx][3 - cotx]^-1 - [x²+1][cscx]² [3 - cotx]^-1 - [x²+1][cosx][cscx][3 - cotx]^-2 [cscx]²

= (2x) [x²+1][cosx][cscx][3 - cotx]^-1 - [x²+1][cscx]² [3 - cotx]^-1 + [x²+1][3 - cotx]^-2[cosx][cscx]³

= [x² + 1][cscx][3 - cotx]^-2 * [(2x)][cosx][3 - cotx] - [cscx][3 - cotx] + [cosx][cscx]²]

Thanks, Melody.......

The original selling price would be given by:

55275/(1-.35) = about 85038.5

There are 5280 ft in one mile

So...   1.3 x 5280  = 6864 ft. per hour

xy = 3/2     solving for y, we have   y =3/(2x)   and substituting this into the other function for y, we have

10x + [3(3/(2x))/5]   = 10x + 9/(10x)  =  10x + (9/10)x^-1

And taking the derivative of this and setting it to 0, we have

10 - (9/10)x^-2  = 0

(9/10)x^-2 = 10

(9/10) = 10x^2

x^2 = 9/100       so...   x = ±(3/10)

Take the second derivative and see if either of the critical points "plugged in" to it produce a positive result...a positive result indicates a minimum.....so we have

10 - (9/10)x^-2  both points plugged into this will = 0

This doesn't tell us anything

Pick two points on either side of -3/10 and see what their slopes are  choosing -1 and -1/2, we have

10 - (9/10)(-1)^-2 = +    and 10 - (9/10)(-1/2)^-2 = -   so   -3/10 is a relative max

Now, pick two points on either side of 3/10 and calculate the slopes  choose (1/5) and 1

10 - (9/10)(1/5)^-2 = -     and   10 - (9/10)(1)^-2   = +   so x =  3/10 is a relative minimum

But,  this function is continuous at everywhere but 0. And at x = 3/10, the fuction has a relative minimum value of 6.And at -3/10, the function has a relative maximum value of -6. And since the function has a relative max at this point, then no absolute minimum exists

We know the opposite side of a triangle (24), an angle (50 degrees), and we are trying to find the hypoteneuse (the length of the wire).....the sine function is what we need...so we have

sin(50) = 24/H    multiply both sides by H and divide both sides by sin(50).....this gives us

H = 24/sin(50) = about 31.33 feet in length

So we want to minimize  2x^4 + 8x^(-4)

Take the derivative and set this to 0

8x^3 - 32x^(-5) = 0       factor

8x^(-5) [x^8 - 4] = 0    ignore the frst factor, it can never be 0

x^8 - 4 = 0

x^8  = 4        take the ±roots....  x = ±(4)^(1/8)  = about ±1.189

Now, take the second derivative and see which critical points (if any) produce a positive result

24x^2 + 160x^(-6)   and both points will produce positive results, so both are minimums

Nice one, Melody....!!!

Let y = (1/3)x^2  = g(x)

We are looking for the point where y = x + c  is tangent to g(x)

Beause the slope of y = x + c  = 1   we are looking for the point where the derivative of g(x) has a slope = 1.

So

g ' (x)  =  2/3x   ....set this = 1    and x = 3/2

And to find the y coordinate of this point we have  y= (1/3)(3/2)^2  = (1/3)(9/4) = 9/12 = 3/4

So, since the point (3/2, 3/4) is also on y = x + c    we have

3/4 = 3/2 + c      subtract 3/2 from both sides

c = -3/4