Consider the function e(h) = ε h h 2 6 m where m is a bound for the third derivative of a function. Show that e(h) has a minimum at p3 3ε/m

Respuesta :

Since , e"(h) = M >0

hence ,  At p3 3ε/M  function e(h) has a minimum of function.

In mathematics, point at which the value of a function is less than or equal to the value at any nearby point (local minimum) or at any point (absolute minimum) or the minimum value of a function is the lowest point of a vertex

e(h) = ∈ /h +( [tex]h^{2}[/tex]/6) M

e'(h) = - ∈ /[tex]h^{2}[/tex] + (2h/6) M

putting , e'(h)  = 0

- ∈ /[tex]h^{2}[/tex] + (2h/6) M  = 0

h = [tex](3 /M)^{1/3}[/tex] *  (∈^(1/3))

e"(h) = 2∈ / [tex]h^{3}[/tex] + (2M)/6                          equation 1

substituting value of h in equation 1, we get

e"(h) = 2∈M / 3∈ + M/3

e"(h) = M >0

hence ,  At p3 3ε/M  function e(h) has a minimum

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