- 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)

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.