## anonymous 5 years ago (a)find y as a function of x, if y'=e^(x-1) and y =1 when x=1. What is the y intercept of this curve? (b)The gradient of a curve is given by y'=e^(2-x) and the curve passes through the point (0,1). What is the equation of this curve? What is its horizontal axis? (c)f=Find y, given that y'=2^(-x), and y=1/2log2 when x =0. (d) It is known that f'(x) = e^x + 1/e and that f(-1)=-1. Find f(0)

1. anonymous

$\int\limits e ^{x-1}$

2. anonymous

$1/e \int\limits e^x d x$

3. anonymous

a) $y = \int\limits_{}^{} e ^{x-1}dx$ u = x-1 du = dx $y=\int\limits_{}^{}e ^{u}du = e ^{u}+C=e ^{x-1}+C$ 1=e^0 +C = 1+C C=0 y = e^x-1

4. anonymous

$1/e[e^x +c]$

5. anonymous

i actually did (a) i need help with the other ones more lol

6. anonymous

b) y = -e^2-x + C 1=-e^2 +C C = 1+e^2 $y = -e ^{2-x} +e ^{2}+1$ c) y=-2^(-x)/ln2 +C 1/2ln2 = -1/ln2 +C C = 3/2ln2 $y = -\frac{1}{\ln 2}(2^{-x} -\frac{3}{2})$ d) f(x) = e^x + x/e +C -1 = e^-1 -1/e +C C = -1 f(x) =e^x +x/e -1 f(0) = 1-1 = 0

7. anonymous

for c is it the same as writing y=1-2^(-x)/log2

8. anonymous

yeah well i get 3/2 or 1.5 instead of 1 but yes you could write it either way

9. anonymous

thats the answer at the back of the book lol

10. anonymous

ahh well book must be wrong..:) is it 1/2*ln2 or 1/(2ln2) for initial conditions

11. anonymous

isnt it the same thing

12. anonymous

no one the log is on top the other its on the bottom

13. anonymous

oh its 1/(2log2)

14. anonymous

thats what i thought i dont know what i did wrong on that one...

15. anonymous

when i did it i got -c=-1/2log2 -1/2log2 c=1/2log2+1/2log2

16. anonymous

theres should be no negative infront of y=-2^(-x)/ln2 +C i think lol

17. anonymous

wait is x=0 or x=1 ? no there needs to be a neg because derivative of -x = -1

18. anonymous

x = 1

19. anonymous

oh sorry i wrote it wrong

20. anonymous

thats it then lol it says 0 above

21. anonymous

yeah im sorry lol i must of did it by accident lol

22. anonymous

it happens, keeps me on my toes :)

23. anonymous

lol thanks heaps you made me learn something new today:)

24. anonymous

your welcome good luck on the rest

25. anonymous

for question (b) when you sub in x = -1 dont you just get -e

26. anonymous

f(x) = e^x + x/e +C when x=-1, = e^-1 + -1/e +C =1/e - 1/e +C =0 + C its not -e because e is in denominator

27. anonymous

oh i was looking at d)...is that right?

28. anonymous

no (b) lol

29. anonymous

ur answer is right but i tried doing it on my own and got -e :(

30. anonymous

ok y=−e^(2−x)+e^2+1 x=-1 =-e^(2-(-1)) +e^2 +1 =-e^3 +e^2 +1 ahh no you cant combine the e's because of the addition

31. anonymous

oh why are you using x=-1

32. anonymous

omg whats wrong with me lol im not reading any of the questions right im really sorry for troubling you for no reason

33. anonymous

haha no prob