Finding maximum value of function

[void lark]

Find the maximum value of this function

[pearl mortar]

do you have any ideas of how you could approach this problem?

[void lark]

i have absolutely no idea . It was asked under the topic of "Application Of Derivatives" though . Tried differentiating and the derivative is even more daunting .

[pearl mortar]

did you find it’s derivative yet?

what information can taking the derivative of a function of the 1st power tell you?

ignore the above—my mistake, sorry. this is transcendental expression, requires some manipulation and idk rn

[void lark]

local maxs and min and about the nature of function . How ever , i dont think that's the method here

[void lark]

its simple actually , but not useful and finding critical points using it is simply impossible for me

[vale dock]

lol, nice problem. Just rewrite it as
3-3(cos(x)-x-√(x²+sin²x))². So, it is obvious that max is 3 if there is such x that cos(x)-x=√(x²+sin²(x)). Since cos(x) is decreasing from 1 to 0 on [0;pi/2] and x+√(x²+sin²(x)) is increasing from 0 on [0;pi/2], such x exists.

[knotty depot]

Don’t u just differentiate and set to zero ?

[vale dock]

no, it is not necessary

[pearl mortar]

i'm curious as to the steps you took to rewrite it as that, i already verified that it matches the given expression but i'm not seeing it. complete the square? an identity?

just not seeing it brah 😔

[vale dock]

If you are asking for the proof, just expand both expressions, and use cos²x+sin²x=1 to see that they are equal.
If you are asking about the way I found it, I first learned the answer from graph, so I knew I have to prove f(x)<=3. Then I made some substitutions (like cos(x)-x=t and cos(x)+x=s) to realize that the whole thing is just (a-b)²>=0 with some specific a and b.

[pearl mortar]

yes, I saw that they are equivalent by that identity, but I didn’t know you used the graph to find the answer. thank you

thought you were working with the expression blind haha, i’ll play around with it later

[void lark]

this is the official solution to this problem that our school published rn

[vale dock]

Lol, it is correct but it is incomplete, because they must also prove that there exists such x that this max=3 is attained, which is equivalent to proving that the equation A-cos(x)=0 has a solution. But they didn't do that.

[void lark]

yeah , that was the DIY part