CPhill

Thanks for that one, Alan.....

Is the fact that we were taking the limit around 0 is what allowed us to use that Maclaurin expansion???? [I think that's what I remember from Cal 2 !!!!]

Undefined.....To see why, note that the sin^-1 (5/4) is asking us to find an angle where the sine is equal to 5/4...but the sine is never greater than 1. So.....we're really being asked to find the sine of an angle that doesn't exist !!!!

If believe you're asking to find the value of the inverse when 21 is substitued into it.... we can do so thusly.....

6(x + 2) - 3= 21      add 3 to both sides

6x + 12 = 24           subtract 12 from both sides

6x = 12                   divide by 6 on both sides

x = 2

Thus (2, 21) is a point on the graph of the original function and (21, 2) is a point on the graph of the inverse.

I can't see any functions......???

Given that f(x) = x2 + 3x + 6 and g(x) = the quantity of x minus three, over two, solve for f(g(x)) when x = 1.

We want to evaluate f(g(1))

So g(x) = (x - 3)/2   and when x =1, g(1) = (1 - 3) /2 = -2/2 = -1

So f(g(1)) = f(-1) = (-1)^2 + 3(-1) + 6 =  1 - 3 + 6 = 4

We want to multiply here.......remember......sin(14) is just a number.......and we're just multiplying this number by 10.

We know the adjacent side (1.5) and the hypoteneuse(5)....we can use the cosne inverse to find the angle

So we have

cos-1(1.5 / 5) = about 72.54°      ...... it is safe......

(1.1*10^3)*(8.6*10^42) = 9.46 x 10⁴⁵

Just type .....6.022 x 10^23

f(x) = [3x + 5]/ 7    first, write f(x) as y...so we have

y = [3x + 5] / 7       we want to get x by itself.....multiply both sides by 7

7y = 3x + 5             subtract 5 from both sides

7y - 5 = 3x              divide both sides by 3

[7y - 5]/ 3  = x        "exchange" x and y  

[7x - 5] /3 = y          for "y,' write f^-1(x)....

[7x - 5] / 3 = f^-1(x)   and that's it !!