Question

for the function f(x)=(x+3)(x-2)^3
a.)find intervals of increase and decrease of f
b.) values of x for which f has a local max and min
c.) find intervals of concavity and inflection points

Answer 1

find the derivative

\[\frac{ d }{ dx } (x+3)(x-2)^{3}\]

can apply the product rule

\[\frac{ d }{ dx }(x+3)(x-2)'^{3}+(x+3)'(x-2)^{3}\]

Then you can take this and set it equal to zero


\[\frac{ d }{ dx } = 3(x-2)^{2}*(x+3)+(x-2)^{3} \]

doing some simplifying

\[ (x-2)^{2} + \frac{ (x-2)^{3} }{ 3(x+3) } = 0 \]

\[ (2-2)^{2} + \frac{ (2-2)^{3} }{ 3(2+3) } = 0 \]

We can see that x = 2 is one of our points where the derivative = 0.

i'm guessing the next step would be to determine whether the function is increasing or decreasing on the particular interval to see whether it's a maximum or minimum.

I guess you could take the value and plug it back into the original function to determine whether it's positive or negative. in a given interval.

Answer 2

give me a minute to tread and udnerstand this, is this for a.) ? @thephysicsman

Answer 3

I guess this could be for B calculating the derivative and setting it equal to zero then finding the points where the derivative = 0. can help us determine whether it's a maximum of a minimum

Answer 4

how do i solve for a.) ?

Answer 5

A plot is attached.

Answer 6

how do i interpret that to find a.) @robtobey

Answer 7

\[\frac{ dy }{ dx }=3\left( x-2 \right)^2\left( x+3 \right)+\left( x-2 \right)^3=\left( x-2 \right)^2\left( 3x+9+\left( x-2 \right) \right)\]
\[=(x-2)^2[4x+7]\]
\[f(x)~is~increasing~if~f \prime(x)>0\]
\[(x-2)^2>0~(always)\]
\[so~f(x)>0,if~4x+7>0 ~or~x>\frac{ -7 }{ 4 }\]

Answer 8

it is decreasing if f'(x)<0
as above if x<-7/4