Question

what are the minimums and maximums of x+sin(x)?

Answer 1

1.find dy/dx
2. find critical points
3.second derivative
then complete

Answer 2

you really don't need the second derivative to find the mins and maxs

Answer 3

I always find the max and mins just with the first derivative rule and look for the sign changes. I thought the second derivative was for finding points of inflections

Answer 4

\[f \prime \left( x \right)=1+\cos x,\]
\[f \prime \prime \left( x \right)=-\cos x\]
\[if~f \prime \prime \left( a \right)>0,then~local~minima.\]
\[if~f \prime \prime \left( a \right)<0,then~local~maxima.\]
\[if~f \prime \prime \left( a \right)=0,then x=a ~is~a~point~of~inflexion.\]

Answer 5

correction\[\frac{ d^2y }{ dx^2 }=-\sin x\]

Answer 6

I thought there was an additional requirement and that is that f'(a)=0

Answer 7

then a is the local max or min as you stated above

Answer 8

so there are not any minimums or maximums?

Answer 9

look at the graph of the function

Answer 10

i did

Answer 11

Due to periodicity, it suffices to look at x, y in (-π, π] (a period).
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Critical points:
f_x = cos x + cos(x+y)
f_y = cos y + cos(x+y).

Set these equal to 0:
cos x + cos(x+y) = 0 and cos y + cos(x+y) = 0.

Subtracting these yields cos x - cos y = 0
==> cos y = cos x
==> y = x + 2πk or y = -x + 2πk for any integer k.

(i) If y = x + 2πk, then we have cos x + cos(2x) = 0 by periodicity
==> cos x + (2 cos^2(x) - 1) = 0
==> 2 cos^2(x) + cos x - 1 = 0
==> (2 cos x - 1)(cos x + 1) = 0
==> cos x = 1/2 or -1.
==> x = ±π/3 + 2πk or x = π + 2πk

With x, y in (-π, π], this yields the critical points (±π/3, ±π/3), (π, π).

(ii) If y = -x + 2πk, then we have cos x + 1 = 0
==> x = π + 2πj for any integer j.

With x, y in (-π, π], this yields the critical point (π, π), which we already have.
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Second Derivative Test:
f_xx = -sin x - sin(x+y), f_yy = -sin y - sin(x+y), f_xy = -sin(x+y).
==> D = (f_xx)(f_yy) - (f_xy)^2 = sin x sin y + (sin x + sin y) sin(x+y).


Since D(π/3, π/3) > 0 and f_xx (π/3, π/3) < 0, we have a local maximum at (π/3, π/3).
Since D(-π/3, -π/3) > 0 and f_xx (-π/3, -π/3) > 0, we have a local minimum at (-π/3, -π/3).

Since D(π, π) = 0, the Second Derivative Test is inapplicable.
However, f(π, π) = 0, and for sufficiently small δ>0, we have f(π+δ, π+δ) < 0 while f(π-δ, π-δ) > 0.
Thus, we have a saddle point at (π, π).
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I hope this helps!