Find the x-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema.

Mathematics

1 answer:

CaHeK987 [17]3 years ago

8 0

Answer:

The function is maxima at x = 0.

Step-by-step explanation:

the function is

$f(x) = x^4 - 8 x^2 + 3$

Differentiate with respect to x.

$f'(x) = 4x^3 - 16 x\\\\Put f'(x) = 0 \\\\4x(x^2-4)=0\\\\4 x(x +2)(x-2) =0\\\\x = 0, - 2 , 2 Now f''(x) = 12 x^2 - 16 \\\\f''(0) = - 16 = negative \\\\f''(-2) = 12(-2)^2 - 16 = 32\\\\f''(2)=12(2)^2 - 16 = 32$

So, the function is maima at x = 0 .

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Find two integers whose sum is -3 and product is -18

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let two integers be x and y

A.T.Q

x+y= -3

or, x= -3-y(i)

and xy= -18

or,(-3-y)y= -18[by(i)]

or,- -3y-y²= -18

or,-y²= -18+3

or, -y²= -15

or, y²=15

therefore y=✓15

from (i)

x= -3-✓15

y=√15

7 0

3 years ago

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