h(x)= x + cos(ax), where a is a positive constant such that 0 < a < 4. For what values of a will h(x) have a relative maximum at x=1?

Question

So I found the derivative and got h'(x) = 1- asin(ax). Then I plugged in 1 for x and set the derivative equal to zero, so I have 1 - asin(a)=0. I think this is right so far, I just don't know how to solve for a at this point. Thank you!!

Answer 1

I made a typo. The derivative is 1+ a sin(ax), sorry.

Answer 2

no, you were right the first time

if y = x + cos(ax) , then
dy/dx = 1 - asin(ax)
so for a max,
asin(ax) = 1
but we are told that x=1
a sin a = 1
sin a = 1/a

very hard equation to solve, if you have a good calculator, set it to radians and start trying some values.
clearly a ≠ 0, or else we are dividing by zero.

how about a = 2
LS = .909.. RS = .5 , LS > RS
a = 3
LS = .14112.. , RS = .333, getting closer to each other , but LS < RS
so a has to be between 2 and 3
a = 2.5
LS = .598 , RS = .4 , LS > RS, so a between 2.3 and 3
a = 2.75
LS = .38.., RS = .36 , LS < RS, so a is between 2.5 and 2.75

Do you get the idea?

I put this into Wolfram and got a = 2.772605
http://www.wolframalpha.com/input/?i=sin+a+%3D+1%2Fa+

Notice that within -3 < a < 3, there are actually 4 answers, and there is symmetry to our curve.
the first positve a = 1.11415714

(use your calculator to show that both of my values work)

looking at larger values of a, shows that there would be an infinite number of solutions for a

http://www.wolframalpha.com/input/?i=sin+a+%3D+1%2Fa+from+0+to+20

Answer 3

just notice that 0 < a < 4,

so our answers would be the two shown in my solution

a = 1.11415714 and
a = 2.772605