Derivatives and graphs f ( x ) = sin ⁡ ( x ) + cos ⁡ ( x ) for 0 ≤ x ≤ 2 π.I have to do the following1) Find the intervals on which f is increasing or decreasing2) Find the local maximum and minimum values of f3) Find the intervals of concavity and the inflections pointsI got until this point when trying to solve problem 1: tan ⁡ ( x ) = 1. The next step in the manual says that x = π / 4 or 5 π / 4. How did they get that x = π / 4 or 5 π / 4?

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Derivatives and graphs  f  (  x  )  =  sin  ⁡  (  x  )  +  cos  ⁡  (  x  ) for   0  ≤  x  ≤  2  π.I have to do the following1) Find the intervals on which f is increasing or decreasing2) Find the local maximum and minimum values of f3) Find the intervals of concavity and the inflections pointsI got until this point when trying to solve problem 1:  tan  ⁡  (  x  )  =  1. The next step in the manual says that   x  =  π      /    4 or   5  π      /    4. How did they get that   x  =  π      /    4 or   5  π      /    4?

trollbabyfeetwa · Accepted Answer

Step 1(1) f is increasing or decreasing accordingly as       f    ′    (  x  )  &amp;gt;  0 or   &amp;lt;  0.Now,       f    ′    (  x  )  =  cos  ⁡  x  −  sin  ⁡  x  =      2    cos  ⁡  (  x  +      π    4    )      f    ′    (  x  )  &amp;gt;  0 if   cos  ⁡  (  x  +      π    4    )  &amp;gt;  0 if   x  +      π    4   lies in the 1st or 4th quadrant.As,   0  ≤  x  ≤  2  π  ,, for       f    ′    (  x  )  &amp;gt;  0  ,  0  ≤  x  &amp;lt;      π    4   or             3      π        2    −      π    4    &amp;lt;  x  ≤  2  π.Similarly for       f    ′    (  x  )  &amp;lt;  0.Step 2(2) For the maxima/minima,       f    ′    (  x  )  =  0    ⟹    tan  ⁡  x  =  1  =  tan  ⁡      π    4      ⟹    x  =  m  π  +      π    4   where m is any integer.      f    ″    (  x  )  =  −  (  sin  ⁡  x  +  cos  ⁡  x  )  =  −      2    cos  ⁡  (  x  −      π    4    )So,       f    ″    (  m  π  +      π    4    )  =  −      2    cos  ⁡  m  π which is   &amp;lt;  0 if m is even   =  2  r (say) where r is any integerSo, the local maximum of f(x) is at   x  =  2  r  π  +      π    4  .As   0  ≤  x  ≤  2  π  ,, for local maximum of   f  (  x  )  ,  x  =      π    4        f          m      a      x        =  f  (      π    4    )  =      2  Similarly, for local minimum.

rkus2zg0 · Accepted Answer

Explanation:We solve for   0  ≤  x  ≤  2  π.   tan  ⁡  x  =  1 if and only if   x  =      tan          −      1        ⁡  1  =      π    4   or   x  =  π  +      tan          −      1        ⁡  1  =            5      π        4  . (      π    4   is the principal value and the tangent is positive in the first and third quadrants.)

Derivatives and graphs. f(x)=sin(x)+cos(x) for 0 leq x leq 2pi.

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