\"Find all the critical numbers of the function g(x) = 4x + sin(4x), without a g

Question

"Find all the critical numbers of the function g(x) = 4x + sin(4x), without a graphing calculator."

When graphing the derivative, g'(x) = 4 + 4cos(4x) on a graphing calculator, it is easy to see that there are multiple critical numbers in a pattern. However when solving this algebraically, I cannot come to the conclusion of multiple critical numbers, only /4. Here is my work:

g'(x) = 4 + 4cos(4x)

0 = 4 + 4cos(4x)

-4 = 4cos(4x)

-1 = cos(4x) , divided both sides by 4

cos-1 (-1) = cos-1(cos(4x)), taking the inverse cosine of both sides

= 4x

x = /4

And yet the true answer is "(2n+1) / 4". Apparently cos-1(-1) yields multiple values, but how do I know that when algebraically solving this? Is this knowledge I'm supposed to have at this point?

Thank you in advance.

Explanation / Answer

g(x) = 4x + sin(4x)

critical points ==> g '(x) = 0

==> 4 + cos(4x) (4) = 0

==> cos(4x) = -1     , here = 4x

cosine function is -1, when = (+/-) , (+/-)( +2) ,(+/-)( +4) , (+/-)( +6) , (+/-)( +8) ---------

==> = (+/-) , (+/-)3 , (+/-)5 , (+/-)7 , (+/-)9 ---------

==> = (2n +1) , n belongs to integers.   {since 1,3,5,7,9---- are in Arithmetical progression. nth term = (2n +1)}

==> 4x = (2n +1)

==> x = (2n +1)/4 , n belongs to integers, are the critical numbers

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