Math question

ItchyArmpit SG

guys i only have one question.. and most of the time i have only one question to ask...

but creating a thread for a question is like a bit "spamming"... so is there any thread that is for u to ask 1 question that i do not know?

anyway the question is... Find the smallest value of the integer b for which
-5x^2 + bx - 2 is negative for all value of x.

ItchyArmpit SG

o one more question... here it goes..

The variables x and y are connected by the equation y = kb^x, where k and b are constants. Experimental values of x and y were obtained. The diagram above (i dunno how to put diagram, but from the words u should be able to visualize) show the straight line graph, passing through the points (0, 1.3) and (11, 0.8), obtained by plotting lg y against x. Estimate

i) the value, to 2 sig. fig, of k and of b,
ii) the value of y when x = 8.

MasterMoogle SG

anyway the question is... Find the smallest value of the integer b for which
-5x^2 + bx - 2 is negative for all value of x.

a = -5 b = b c = -2

b^2- 4ac<0

b^2-4(-5)(-2)<0

b^2 - 40<0

b^2<40

So smallest value = -6, since they want integer value

MasterMoogle SG

lg k = y-axis intercept and lgb = gradient

just see from graph for part 2

ItchyArmpit SG

a = -5 b = b c = -2

b^2- 4ac<0

b^2-4(-5)(-2)<0

b^2 - 40<0

b^2<40

So smallest value = -6, since they want integer value

wow... actually i did the same working, just the last part i dont get it... cuz b is smaller than something... is infinite what... but now i get it, cuz is square.. anything also become positive... so -6 is the smallest.. :)

ItchyArmpit SG

lg k = y-axis intercept and lgb = gradient

just see from graph for part 2

sorry but i dont get it... how do u get the lg k = y axis intercept and lg b = gradient?

thanks for teaching! :)

MasterMoogle SG

sorry but i dont get it... how do u get the lg k = y axis intercept and lg b = gradient?

thanks for teaching! :)

y = kb^x

lgy = lg(kb^x)

lgy = lgk + lgb^x

lgy = lgk + xlgb

lgy = (lgb)x + lgk

Y = mX + c, where m is gradient and c is y-axis intercept

Y now is lgy, X is x, so m = gradient = lgb and c = y-axis intercept = lgk.

tada!

ItchyArmpit SG

why didnt i think of it... THANKS!!! big thanks!

u helped a lot :)

ItchyArmpit SG

hi i got question again... here it goes

g(x) = 3 - l2x - 4l is defined for the domain 0 <= x <= 5.

a) Sketch the graph of g(x) and state the range of g corresponding to this domain.

for this question, i already sketch, and i am sure i am correct... the only thing is that i got -1 <= x <= 3 for the domain... the answer shows -3 <= x <= 3 ... so what is your answer? and how u get it? no need to sketch the graph out.. cuz i think i got it... am i right?

b) have to do a) first... and if i got a) i think i know how to do this :)

ItchyArmpit SG

and one more question... about the kinematics in a maths o lvl..

we always encounter the type of question that goes "find t when the particle is instantaneously at rest" so we go v = 0 and solve the quadratic equation...

but i saw one question that says velocity is MAXIMUM... not at rest.. what do i do?
do i differentiate and equate to 0? since this is the way to find maximum or minimum thing.. urmm i lost that question already... just that it came to my mind suddenly :)

thanks!

ItchyArmpit SG

ah another question lol!

a) Sketch the graph of y = ln(squareroot 2x - 1) for x > 0.5

b) Determine the equation of the straight line which would need to be drawn on the graph of y = ln(squareroot 2x - 1) in order to obtain the graphical solution of the equation e^2x + e^4 = 2xe^4

i am specifically asking about the part b), what does it mean, i dun really understand the question... and how to do? it is alright if u explain in words, and not do working.. any help will be appreciated, thanks!

wee_ws SG

hi i got question again... here it goes

g(x) = 3 - l2x - 4l is defined for the domain 0 <= x <= 5.

a) Sketch the graph of g(x) and state the range of g corresponding to this domain.

Hi,

We start by removing the modulus sign.

|2x - 4| = 2x - 4 when 2x - 4 >= 0, i.e. x >= 2. So, when x >= 2, g(x) = 3 - (2x - 4) = 7 - 2x.

|2x - 4| = -(2x - 4) when x < 2. So, when x < 2, g(x) = 3 + (2x - 4) = 2x - 1.

Now, we sketch g(x) = 2x - 1 for 0 <= x < 2

and g(x) = 7 - 2x for 2 <= x <= 5

in order to find the range within the given domain.

Thanks!

Cheers,
Wen Shih

wee_ws SG

but i saw one question that says velocity is MAXIMUM... not at rest.. what do i do?
do i differentiate and equate to 0? since this is the way to find maximum or minimum thing.. urmm i lost that question already... just that it came to my mind suddenly :)

thanks!

Hi,

To find maximum v, we need to set dv/dt = 0. Thanks!

Cheers,
Wen Shih

wee_ws SG

b) Determine the equation of the straight line which would need to be drawn on the graph of y = ln(squareroot 2x - 1) in order to obtain the graphical solution of the equation e^2x + e^4 = 2xe^4

i am specifically asking about the part b), what does it mean, i dun really understand the question... and how to do? it is alright if u explain in words, and not do working.. any help will be appreciated, thanks!

Hi,

It means that given e^2x + e^4 = 2xe^4, could we manipulate it so that one side will look like ln {sqrt(2x) - 1} and another side will look like the equation of a straight line to be drawn.

Do try it!

Thanks!

Cheers,
Wen Shih

ItchyArmpit SG

Hi,

It means that given e^2x + e^4 = 2xe^4, could we manipulate it so that one side will look like ln {sqrt(2x) - 1} and another side will look like the equation of a straight line to be drawn.

Do try it!

Thanks!

Cheers,
Wen Shih

haha now u mention, i rmb i do this type of question b4.. ah how silly i am...

anyway here goes my working...

e^2x + e^4 = 2xe^4
e^2x = 2xe^4 - e^4
e^2x = e^4 (2x - 1)
ln both side
2x = lne^4 + ln(2x-1)
2x = 4 + ln(2x-1)

then i got stuck here...

then i tried working again...

e^2x + e^4 = 2xe^4
e^2x = 2xe^4 - e^4
e^2x = e^4 (2x - 1)
square root everything (is this "legal"?)
e^x = e^2 (square root 2x - 1)
ln both side
x = 2 + ln(squareroot 2x -1)
ln (squareroot 2x-1) = x - 2

my question is... the 2nd working i did, the squareroot everything, issit legal? if yes did i do it correctly? is i did, then the answer is y = x - 2?

thanks!

ItchyArmpit SG

Hi,

We start by removing the modulus sign.

|2x - 4| = 2x - 4 when 2x - 4 >= 0, i.e. x >= 2. So, when x >= 2, g(x) = 3 - (2x - 4) = 7 - 2x.

|2x - 4| = -(2x - 4) when x < 2. So, when x < 2, g(x) = 3 + (2x - 4) = 2x - 1.

Now, we sketch g(x) = 2x - 1 for 0 <= x < 2

and g(x) = 7 - 2x for 2 <= x <= 5

in order to find the range within the given domain.

Thanks!

Cheers,
Wen Shih

urmm thanks for ur help, but i dont really understand... i only understand the ones i highlighted green... can u explain more? and as i said earlier i had worked out an answer, can u please tell me if my answer is correct? the answer is -1<= x <= 3

sorry to trouble u so much haha.

and thanks for the kinematics thing :) i understand.

oh oh... i forgot to say that..

from the earlier question, the ln(squareroot 2x-1) thing, i get it le... and i know i am correct, cuz from the 1st working i did, i can divide by 2 the whole thing, and 0.5 ln(2x - 1) is equalz to ln(squareroot 2x - 1)... :) so this shows that my 2nd working de step are all "legal"...

thanks :)

ItchyArmpit SG

ah haha.. got one more thing i wanna confirm... can someone show me the steps to differentiate (cos x)^3 - 3 (cos x)

thanks!

ItchyArmpit SG

ah haha.. got one more thing i wanna confirm... can someone show me the steps to differentiate (cos x)^3 - 3 (cos x)

thanks!

ok nvm i got it le.... for those who are curious, u differentiate cosx square then use the result and differentiate again with ( cos x ) (cos x) ^2

wee_ws SG

can u explain more? and as i said earlier i had worked out an answer, can u please tell me if my answer is correct? the answer is -1<= x <= 3

Hi,

Between 0<= x < 2, sketch g(x) = 2x - 1; between 2 <= x <= 5, sketch g(x) = 7 - 2x.

The correct answer is -3 <= g(x) <= 3. The value of -3 lies on g(x) = 7 - 2x when x = 5 and the value of 3 is the location where both graphs meet when x = 2.

Do take note that the range is always referring to the set of values on y-axis or g(x)- axis in this case.

Thanks!

Cheers,
Wen Shih

P.S. I have written a short article about this problem at

http://weews.wetpaint.com/page/Sketching+modulus+graphs