Question

How can you say that the function is maximum or minimum? i have a solution then check if its correct....

Answer 1

y=2+12x-x^3
=12-3x^2
=3(x^2-4)
= (x+2)(x-2)
x+2=0 x-2=0
x=-2 x=2
a=2
x,2;x=1
y=(x+2)(x-2)
=(3)(-1)
y′
=-3

x.2;x=3
y=(3=2)(3-2)
y′
=5

x=a;x=2
y=(2+2)(2-2)
y′
=0
so x=2 at minimum?

Answer 2

min is at x= -2

max is at x = 2

Answer 3

@ jim..can you explain how did you come up with that?pls

Answer 4

y=2+12x-x^3


y'=-3x^2+12


y''=-6x



Set first derivative equal to zero and solve to get


0 = -3x^2+12


-12 = -3x^2


4 = x^2


x^2 = 4


x = 2 or x = -2


So we know that at those x values, the tangent line is horizontal, therefore, those points are either minimums or maximums



To figure out which is which, plug in each extrema into the second derivative


y'' = -6x


so when x = 2, y'' = -6(2) = -12


Since this is less than zero, this means that we have a max here


when x = -2, y'' = -6(-2) = 12


because this is greater than zero, we have a min here

Answer 5

thanks jim

Answer 6

which part doesn't make sense?

Answer 7

your using 2nd derivative test right?while i'm using 1st derivative test.but the answer will be the same?

Answer 8

yes either method will get you the same answer, but the first derivative test takes a bit more work

Answer 9

with the first derivative test, you have to examine the change in sign of the derivatives

Answer 10

ok tnx.i'll try to use the 2nd...

Answer 11

whereas with the second derivative test, all you need is the sign itself

Answer 12

things to remember

if f''(x) < 0, then the function is concave down at that point

if f''(x) > 0, then the function is concave up at that point

Answer 13

@ jim..can you explain how did you come up with that?pls

Answer 14

tnx again...