17339884
| Question | Answer |
| sinhx graph: |
Image:
Sinh Plot (image/gif)
|
| coshx graph: |
Image:
Math4 6f 3 (image/png)
|
| coshx crosses the y-axis at.... | (0,1) |
| The shape of coshx is called a.... | Catenary |
| tanhx graph: |
Image:
Tanh Real (image/gif)
|
| Asymptotes of tanhx : | y=1 and y=-1 |
| Exponentially define coshx | coshx = (e^x + e^-x)/2 |
| Exponentially define sinhx | sinhx = (e^x - e^-x)/2 |
| Exponentially define tanhx | (e^x - e^-x)/(e^x + e^-x) (as tanhx = sinhx / coshx) |
| If we differentiate sinhx and coshx what do we get? (Note we dont need to know this year) | y=sinhx , dy/dx=coshx y=coshx , dy/dx=sinhx |
| sinhx-1 , coshx-1 , tanhx-1 can be also stated like... | arsinhx , arcoshx , artanhx |
| sinhx-1 graph: |
Image:
Graph Arsinh (image/gif)
|
| coshx-1 graph: |
Image:
Graph Arcosh (image/gif)
|
| We only show 1 half of the coshx-1 graph, meaning coshx-1(4) gives the values..... | +/- 2.1 |
| tanhx-1 graph: | |
| tanhx-1 has the asymptotes..... | x=1 and x=-1 |
| What is the method we use for defining y=arsinhx exponentially? | Multiply both sides by sinhx: sinhy=x Replace with : (e^y - e^-y)/2 = x , then (e^y - e^-y) = 2x. Multiply everything by e^y to remove "e^-y" and make into quadratic: (e^y) - 2xe^y - 1 = 0. To solve use quadratic formula: e^y = 2x+/-root(4x^2-4(1)(-1)) / 2. This cancels to e^y=x+/-root(x^2+1)) , which goes to y=ln(x+/-root(x^2+1)). Now we know root(x^2+1)>0 and the ln x function only works when x>0 , so arsinhx = ln(x+root(x^2+1)) |
| Note on how e^y = 2x+/-root(4x^2-4(1)(-1)) / 2 cancels to e^y=x+/-root(x^2+1)) : | root(4x^2+4) = root(4(x^2+1)) Same as root(4) x root(x^2+1) Same as 2root(x^2+1) whcih divides by to to get 1, so essentially it disappears. We also divide the 2x at the start to get: e^y=x+/-root(x^2+1)) |
| What is the method we use for defining y=arcoshx exponentially? | Multiply both sides by coshx: coshy=x Rewrite as exponential form: e^y+e^-y/2 =x Multiply everything by e^y and form quadratic again and solve with quadratic formula: arcoshx=ln(x+/-root(x^2-1)). We know though that inverse functions must have only 1 solution so we can ignore the negative branch, so: arcoshx=ln(x+root(x^2-1) when x≥1 |
| What is the method we use for defining y=artanhx exponentially? | Multiply both sides by tanhx: tanhy=x Rewrite as exponential form: e^y-e^-y/e^y+e^-y =x . Multiply by e^y and rearrange for e^y on 1 side: (e^y)^2 - x(e^y)^2 = x+1. Factor out e^y : (e^y)^2 [1-x] = x+1 , rearrange for: e^2y = 1+x/1-x . Log both sides: 2y = ln(1+x/1-x). So artanhx = (ln(1+x/1-x)/2) Only works for -1<x<1 (or |x|<1 ) |
| Write sinhx = 5 into exact form | x=sinx-1(5) , ln(5+root(5^2+1)) So ln(5+root(26)) |
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