Question

If f'(x)=(x+2)^2*(x-4). what would the value of x be where f has a local max. or min. and how would I find the point of inflection?

Answer 1

equate (x+2)^2*(x-4)=0
and solve

Answer 2

don't forget to use the second derivative

Answer 3

what would the second derivative be

Answer 4

f'(x)= (x+2)^2*(x-4)=0
f'' (x)= product rule and chain rule cuz of this \[ (x+2)^2*(x-4)\]

Answer 5

now u know how to solve it

Answer 6

so f"(x)= 2(2x^2 - 2x) ?

Answer 7

@Goaliess1 almost correct 2(x+2)( 1)(x-4)+ (1)(x+2)^2

Answer 8

would it be 3x^2 -12

Answer 9

find x

Answer 10

the question is no over yet

Answer 11

we have x=2

Answer 12

so would the local max be at x=2 and there is no local min?
also would the points of inflection be x=+2 and -2?

Answer 13

this shows first derivative f'(x)=(x+2)^2*(x-4)

Answer 14

yes but can't you find the point of inflection by the change in concavity?

Answer 15

(-2, 4) critical pts

Answer 16

f(-2)=2(x+2)( 1)(x-4)+ (1)(x+2)^2
f(4)= 2(x+2)( 1)(x-4)+ (1)(x+2)^2
this is the inflection point will decided whether this eq is local min or max

Answer 17

sorry about that got little bit confuse, really need to use the second derivative to find the point of inflection

Answer 18

what @matricked is right
f'(x)=(x+2)^2*(x-4)
x=-2
x=4

Answer 19

about the f(x) used it to find the global max and min

Answer 20

sorry about early

Answer 21

just finish solving this f(-2)=2(x+2)( 1)(x-4)+ (1)(x+2)^2
f(4)= 2(x+2)( 1)(x-4)+ (1)(x+2)^2 and finding if is loca max or min then u are done
have awonderful day im off now