Have taken any calculus yet?A function is given by f (x)= ax^2-6x+ b sin (3x) where a and b are real constants. Examine which values of a and b are possible
if the function has a maximum when x=0
To receive help you need to inform us where you are stuck, the method you learned in your class, etc. This problem can be done using calculus but we do not know if you have taken calculus. With a little thought it probably can be done using algebra. Since you did not give us any information we have no idea how to help you.A function is given by f (x)= ax^2-6x+ b sin (3x) where a and b are real constants. Examine which values of a and b are possible
if the function has a maximum when x=0
thank you for your concern.To receive help you need to inform us where you are stuck, the method you learned in your class, etc. This problem can be done using calculus but we do not know if you have taken calculus. If a little thought it probably can be done using algebra. Since you did not give us any information we have no idea how to help you.
I wholeheartedly agree that we need to know your mathematical background to give you proper help, but I point out thatA function is given by f (x)= ax^2-6x+ b sin (3x) where a and b are real constants. Examine which values of a and b are possible
if the function has a maximum when x=0
Hint:thank you for your concern.
I have taken calculus and I need tips how to solve this problem
Can you find the derivative of the function? Please show us your work and we'll proceed from there.any news ?
Additional condition to satisfy (for local maximum)d/dx(ax^2-6x+bsin(3x)=2ax-6+3bcos(3x)
since it has a maximum when x=0, Then
6+3bcos(0)=0; 3b=-6
b= -2
this What I have Done so far.....
For maxima at x = 0,Mr. Khan
I did not get your point regarding the 2nd derivative of the function. I have found the 2nd derivative.
Which values of a and b are possible?
There is no chance for your answer to be correct (Sorry). The reason is the format of your answer. First let's say you said a<= (-7,4]. This means a is less that every number in the interval between -7 up to and including 4. This means that a<=-7. Do you see this???Since b= -2 and for maxima at x=0
f”(0)=2a+18sin(3x)<=0
2a<= -18sin(3x)
a<=-9sin(3x)
since -1<= sin(3x)<=1
a<=( - inf.;9]
Is it correct?
f”(0)=2a+18sin(3x)<=0Since b= -2 and for maxima at x=0
f”(0)=2a+18sin(3x)<=0
2a<= -18sin(3x)................................................Incorrect
a<=-9sin(3x)
since -1<= sin(3x)<=1
a<=( - inf.;9]
Is it correct?
The attachment is very small. Can you please post a larger one? Thanks.I have attached my soluation up...would you pls check