# Math Help - Can someone please check my homework?

1. ## Can someone please check my homework?

Hi could one of the geniuses out there please check my homework?

Find the stationary points for f(x)=-2x^3+27x^2-84x+32
=-2(3x^2)+27(2x)-84(1)
=-6(x^2-9x+14)
=-6(x-2) (x-7)
Stationary points x=2 x=7

Then classify the left hand side using the First Derivative Test
x=2 xl=1 xr=3
f'(1)=-6(1) (-32) =190 >0
f'(3)= -6 (3) (-32) =-576 <0
So f has a local minimum at x=2

Classify the right hand side using the Second Derivative Test
x=7
Second derivative
-6(2x) + 54
=-12x + 54
=-6(2x - 9)
For stationary point x=7 xl=6 xr=8
f"(6) =-6((2x6)-(9x6))=252>0
f"(8) = -6((2x8)-(9x8)=336>0

Find the y coordinates x=2 x=7
f(2)= -16+1458-168+32=1306
f(7)=-56+5103-588+32 = 4491

I know theres a lot but would be greatful for any help.

2. Hello, stewpot!

I don't understand the procedure you're using . . .
. . First Derivative Test for the "left side"?
. . Second Derivative Test for the "right side"?

Who told you to do it that way ??

Find and identify the stationary points for: . $f(x)\:=\:-2x^3+27x^2-84x+32$
Set the first derivative equal to zero: . $f'(x) \:=\:-6x^2 + 54x - 84 \:=\:0$

. . which factors:. . $-6(x - 2)(x - 7) \:=\:0$

. . and has roots: . $x \:=\:2,\:7$

Stationary points: . $(2,\text{-}44),\;(7,81)$

Second Derivative Test: . $f''(x) \:=\:-12x + 54 \:=\:-6(2x-9)$

. . $f''(2) \:=\:-6(4-9) \:=\:+30\;\hdots\text{ concave up: }\cup\;\hdots\text{ minimum at }(2,\text{-}44)$

. . $f''(7) \:=\:-6(14-9) \:=\:\text{-}30\;\hdots\text{ concave down: }\cap\;\hdots\text{ maximum at }(7,81)$

3. ## Can you please show me how you arrived at this!

H Soroban,

Thanks for you speedy answer, I'm a little puzzled, can you break down how you arrived at the coordinates from [IMG]file:///Users/andystewart/Library/Caches/TemporaryItems/moz-screenshot.jpg[/IMG]-6(x-2)(x-7)=0.

I've tried most solutions and keep comming up with different coordinates!

Many thanks

Stewpot