Question

(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)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

for c is it the same as writing

y=1-2^(-x)/log2

Answer 8

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

Answer 9

thats the answer at the back of the book lol

Answer 10

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

Answer 11

isnt it the same thing

Answer 12

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

Answer 13

oh its 1/(2log2)

Answer 14

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

Answer 15

when i did it i got

-c=-1/2log2 -1/2log2

c=1/2log2+1/2log2

Answer 16

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

i think lol

Answer 17

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

Answer 18

x = 1

Answer 19

oh sorry i wrote it wrong

Answer 20

thats it then lol
it says 0 above

Answer 21

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

Answer 22

it happens, keeps me on my toes :)

Answer 23

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

Answer 24

your welcome
good luck on the rest

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

no (b) lol

Answer 29

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

Answer 30

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

Answer 31

oh why are you using x=-1

Answer 32

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

Answer 33

haha no prob