anonymous
  • anonymous
(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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits e ^{x-1}\]
anonymous
  • anonymous
\[1/e \int\limits e^x d x\]
dumbcow
  • dumbcow
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

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anonymous
  • anonymous
\[1/e[e^x +c]\]
anonymous
  • anonymous
i actually did (a) i need help with the other ones more lol
dumbcow
  • dumbcow
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
anonymous
  • anonymous
for c is it the same as writing y=1-2^(-x)/log2
dumbcow
  • dumbcow
yeah well i get 3/2 or 1.5 instead of 1 but yes you could write it either way
anonymous
  • anonymous
thats the answer at the back of the book lol
dumbcow
  • dumbcow
ahh well book must be wrong..:) is it 1/2*ln2 or 1/(2ln2) for initial conditions
anonymous
  • anonymous
isnt it the same thing
dumbcow
  • dumbcow
no one the log is on top the other its on the bottom
anonymous
  • anonymous
oh its 1/(2log2)
dumbcow
  • dumbcow
thats what i thought i dont know what i did wrong on that one...
anonymous
  • anonymous
when i did it i got -c=-1/2log2 -1/2log2 c=1/2log2+1/2log2
anonymous
  • anonymous
theres should be no negative infront of y=-2^(-x)/ln2 +C i think lol
dumbcow
  • dumbcow
wait is x=0 or x=1 ? no there needs to be a neg because derivative of -x = -1
anonymous
  • anonymous
x = 1
anonymous
  • anonymous
oh sorry i wrote it wrong
dumbcow
  • dumbcow
thats it then lol it says 0 above
anonymous
  • anonymous
yeah im sorry lol i must of did it by accident lol
dumbcow
  • dumbcow
it happens, keeps me on my toes :)
anonymous
  • anonymous
lol thanks heaps you made me learn something new today:)
dumbcow
  • dumbcow
your welcome good luck on the rest
anonymous
  • anonymous
for question (b) when you sub in x = -1 dont you just get -e
dumbcow
  • dumbcow
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
dumbcow
  • dumbcow
oh i was looking at d)...is that right?
anonymous
  • anonymous
no (b) lol
anonymous
  • anonymous
ur answer is right but i tried doing it on my own and got -e :(
dumbcow
  • dumbcow
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
dumbcow
  • dumbcow
oh why are you using x=-1
anonymous
  • anonymous
omg whats wrong with me lol im not reading any of the questions right im really sorry for troubling you for no reason
dumbcow
  • dumbcow
haha no prob

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