f(x) g(x)
Do you mean the following functions ? \[f(x)=-4\left( x-6 \right)^{2}+3\] and \[g(x)=2\cos \left( 2x-\pi \right)+4\]
yes @marinos
The maximum of cosine is 1, then g max is 2+4=2 The the vertex of the flipped parabola is when is 0, cause it's negative otherwise. So f max is 0+3 =3
can you explain how @Ahmad-nedal
Remember that a square is always non-negative and that the absolute value of (sine and) cosine function(s) is at most 1.
But the thing is, these two values are not satisfied at the same x coordinate, I think you will now be needing calculus 1 to find that point where x gets you the maximum
What is your grade right now @jacey.stewart ?
You use equation x=-b/2a to find the maximum Im just confused on the cosine part
B
The maximum of any sinsouidal function (since, cosx) is 1, therefore the maximum of cosine is 2
We have that \[\left( x-6 \right)^{2}\ge0\] since it is a square. Therefore \[-4\left( x-6 \right)^{2}\le0\] so \[f(x)=-4\left( x-6 \right)^{2}+3\le3\] implying that the maximum value of the function \[f(x)\] is 3. Can you argue similarly for the other function ?
In my opinion, I think the question is not well written. Again if the maximum of f and g are achieved in two different x coordinates. Thus you cannot determine which point gives you the maximum MULTIPLE OF F(X) AND G(X)
I repeat, you may need deffrentiation to find the maximum of f times g
Is what I'm saying make sense jacy?
@Ahmad-nedal No differentiation is needed for these functions. There maximum value can be obtained using inqualities.
I think it ultimately asked for the maximum of the multiple of F and G, isn't she?
Since \[\cos \left( 2x-\pi \right)\le1\] we have \[2\cos \left( 2x-\pi \right)\le2\] so \[g(x)=2\cos \left( 2x-\pi \right)+4\le2+4=6\] therefore the maximum value of this function is 6, and it is the greatest of the maximum values of the two initial functions.
@Ahmad-nedal I didn't see any reference to the product of f(x) and g(x) (in which case you do need differentiation to find the max value)
Then you are right, it will be kinda trevial if considered two separate questions Thanksfor clarification
You are welcome.
Join our real-time social learning platform and learn together with your friends!