y = x + (1/x)

y = x + x^-1

dydx = 1 – x^-2

Set to 0 and solve

0 = 1 – x^-2

mult everything by x^2

0 = x^2 – 1

1 = x^2

x = 1 and x = -1

take second derivative:

dydx = 1 – x^-2

d2ydx2 = – (-2) x^-3

d2ydx2 = 2 x^-3

when x=1 second derivative is positive, so it concaves up, so x=1 is a local min

when x=-1 second derivative is negative, so it concaves down, so x=-1 is a local max

sub in to get optimal values

y = 1 + 1/1 = 2, so (1, 2)

y = -1 + 1/-1 = -2, so (1, -2)

So, at x=1, a local min of 2 occurs

So, at x=-1, a local max of -2 occurs

There are no global min or max in this one.

Take a look using a function grapher:

http://www.walterzorn.com/grapher/grapher_e.htm

for example you can buy items using USD, Pounds in india but local currency is Indian Rupee.

another example is you can use USD, Euros, British Pounds, Swiss Franc in whole europe but if you talk in terms of local currency then for UK its pound, for switzerland it’s swiss franc for America it’s USD and for europe other then mention countries their local currency is Euros.

hope you understood the answer to your question.

]]>we set this equal to 0

18x – 18x^2 = 0

18x(1-x) = 0

the critical points are at x = 0 and x = 1

the second derivative is

f ” (x) = 18 – 36x

f ” ( 0 ) = 18 > 0, so f has a local min at x = 0

f ” (1) = 18 – 36 < 0, so f has a local max at x = 1

so at x = 1, the local maximum value is f(1) = 8 + 9 – 6 = 11

and at x = 0 , the local minimum value is f(0) = 8.

f(x) = 8 + 9x^2 − 6x^3 ]]>

y = x/(x^2+4)

y’ = (4 – x^2)/(x^2 + 4)^2 = 0

x^2 = 4

x = -2 , y = -1/4 =>Global minima

x = 2 , y = 1/4 => Global Maxima

there are no local extrema

2) y = x^5 – 5x + 4

y’ = 5x^4 – 5 = 0

x = – 1 , y = 8 => local maxima

x = 1 , y = 0 => local minima

second derivative test:

y” = 20x^3

y”(-1) = -20 < 0 => hence a max

y”(1) = 20 > 0 => hence a min

2) find the local extrema of f(x)=(x^5)-5x+4 using the second derivative test

]]>Taken from a website:

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