(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)

- anonymous

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- anonymous

\[\int\limits e ^{x-1}\]

- anonymous

\[1/e \int\limits e^x d x\]

- 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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## More answers

- anonymous

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

- anonymous

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

- 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

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

- dumbcow

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

- anonymous

thats the answer at the back of the book lol

- dumbcow

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

- anonymous

isnt it the same thing

- dumbcow

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

- anonymous

oh its 1/(2log2)

- dumbcow

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

- anonymous

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

- anonymous

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

- dumbcow

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

- anonymous

x = 1

- anonymous

oh sorry i wrote it wrong

- dumbcow

thats it then lol
it says 0 above

- anonymous

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

- dumbcow

it happens, keeps me on my toes :)

- anonymous

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

- dumbcow

your welcome
good luck on the rest

- anonymous

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

- 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

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

- anonymous

no (b) lol

- anonymous

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

- 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

oh why are you using x=-1

- 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

haha no prob

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