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First derivative test
First derivative test: classify the critical points based on change of the sign of f'(a)
f′(x) > 0 for x < a and f′(x) < 0 for x > a, f′(x) < 0 for x < a and f′(x) > 0 for x > a, no change.
The gradient is used to find the extreme point of a function of several variable in a manner similar to the 1 dim case.
Theorem (First derivative test in dim n). Suppose a is a local extreme point of a differentiable function f. Then
(1) f′(a) = 0, (2) ∇f(a) = 0, (3) ∂f / ∂xi (a) = 0 for all i = 1, ..., n.
Exercise. Let f( x1, x2 ) = x12 + x22. Find the extrema. Solve the equations:
∂f / ∂x1 = 2x1 = 0 ⇒ x1 = 0, ∂f / ∂x2 = 2x2 = 0 ⇒ x2 = 0.
Hence the point ( 0, 0 ) is the only critical point of f (minimum).
Exercise. Let
g( x1, x2 ) = x12 - x22.
Then the calculation is the same as above, and we obtain ( 0, 0 ) as the only critical point (saddle).
Exercise. Let
f = ( x1, x2 ) = x12.
Then
∂f / ∂x1 = 2x1 = 0 ⇒ x1 = 0, ∂f / ∂x2 = 0.
Hence the critical points are
{ ( 0, x2 ) : x2 ∈ ℝ },
which form a line.
Proof of the theorem. Suppose the point a is a maximum for f: ℝn → ℝ. Suppose a = ( a1, ..., an ). Consider
f1: ℝ → ℝ, f1(x) = f( x1, a2, ..., an ),
which is differentiable. Then x = a1 is a max of f1. Now, by the First Derivative Test from Calc 1, we have
f1′( a1 ) = 0.
This derivative is equal to the partial derivative:
∂f / ∂x1 (a) = 0.
Continue with fk: ℝ → ℝ,
fk(x) = f( a1 ... ak-1 × ak+1 ... an).
Then fk is differentiable, and x = ak is a maximum. Find
f′k (ak) = 0,
so
∂f / ∂xk (a) = 0.
Thus all partial derivatives of f are equal to 0. Hence the gradient is zero too:
∇f(a) = 0 (a vector),
so
f′(a) = 0 (a linear map). QED