Find the possible values of constants a and b in a quadratic function

Mahgoub

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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
 
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
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.
 
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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.
thank you for your concern.
I have taken calculus and I need tips how to solve this problem
 
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
I wholeheartedly agree that we need to know your mathematical background to give you proper help, but I point out that

\(\displaystyle ax^2 - 6x + b * sin(3x)\) is NOT a quadratic function.

A quadratic function has three terms, one of which is a constant. None of the three terms in the specified function is a constant.

An important clue if you do not know calculus is that

\(\displaystyle 0 \le |\ b * sin(3x)\ | \le |\ b\ | \ \because \ 0 \le |\ sin(3x)\ | \le 1.\)
 
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thank you for your concern.
I have taken calculus and I need tips how to solve this problem
Hint:

What is the value of the derivative of the function at the local maxima/minima of the function?
 
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.....
 
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.....
Additional condition to satisfy (for local maximum)

\(\displaystyle \dfrac{d^2y}{dx^2}\) <= 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?
 
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?
For maxima at x = 0,

\(\displaystyle \dfrac{d^2}{dx^2}f(0)\) ≤ 0

so now think what must be the domain of 'a' to have f(0) maximum?
 
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?
 
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?
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???

Now onto your answer. a<=( - inf.;9] says that a<= -inf. But neg infinity is Not a number and anything less than neg infinity is not a number. So you should have said after writing a<=( - inf.;9] that there is no such a.

The good news is that getting to a<=( - inf.;9] was wrong. You have the wrong 2nd derivative. Please try again.
 
Since 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?
f”(0)=2a+18sin(3x)<=0

2a+18bsin(3x)<=0

2a <=0 ............................because we have shown that b = 0 for extrema = 0

a <=0
 
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