Finding local max and min

[kggirl]

The derivative of a function is expressed by f' = (x-1)^3(x+2)(x-3). Find all the points where f has a. a local minimum b. a local maximum

 

[pka]

You have posted three different questions with no work shown.
What have YOU done on anyone of them?
Show us where you are having difficulty!

 

[stapel]

Hint: What is the relationship between the first derivative of a function and the local max/min points of that function?

Eliz.

 

[mammothrob]

one of the interpretations of the first derivative is slope.
To find min and max values you need find the critical numbers of f'(x)
which happens to be the zeros. Your derivative is nicely factored already so it shouldnt be that tough to do.
what makes f'(x) = 0?
its nice to have the original function to test your Critical Numbers to describe min/max values but the f''(x) will give you the + - concavity values of the function.

 

[kggirl]

You have posted three different questions with no work shown.
What have YOU done on anyone of them?
Show us where you are having difficulty!

The derivative of a function is expressed by f' = (x-1)^3(x+2)(x-3). Find all the points where f has a. a local minimum b. a local maximum

This is what I have:

f' = 0 at x=1, x=-2 and x = 3

if x =0 then (0-1)^3(0+2)(0-3) = 0 (positive and goes up)
if x = 1.5 then (1.5-1)^3(1.5+2)(1.5-3) = -1.31 (negative and goes down)
if x = 5 then (5-1)^3(5+2)(5-3) = 56 (positive and goes up again)

so would the local max be 5 and the local min be 0?

 

[soroban]

Hello, kggirl!

The derivative of a function is expressed by \(\displaystyle f '(x)\:=\x\,-\,1)^3(x\,+\,2)(x\,-\,3)\).
Find all the points where $\displaystyle f$ has:
a) a local minimum . . b) a local maximum

This is what I have:

$\displaystyle f'(x)\,=\,0$ at $\displaystyle x\,=\,1,\,-2,\,3$ . . . . correct!

$\displaystyle f'($-$\displaystyle 3)\,=\,($-$\displaystyle 3-1)^3($-$\displaystyle 3+2)($-$\displaystyle 3-3)\,=\,-384$ . negative $\displaystyle \searrow$ . . . . you forgot this one

$\displaystyle f'(0)\,=\,(0-1)^3(0+2)(0-3)\,=\,6$ . positive $\displaystyle \nearrow$

$\displaystyle f'(2)\,=\,(2-1)^3(2+2)(2-3)\,=\,-4$ . negative $\displaystyle \searrow$

$\displaystyle f'(4)\,=\,(4-1)^3(4+2)(4-3)\,=\,162$ . positive $\displaystyle \nearrow$

Your work is correct . . . I just changed some numbers.

Then make a sketch!


                            |   -----
          \                 /           \           /
              -----         |               -----
      - - + - - + - - + - - + - - + - - + - - + - - +
         -3    -2           0     1     2     3     4

Now you can SEE where the maximums and minimums are . . .

 

[kggirl]

thank you so much!

thank you