Question

what is the local maximum value of the function g(x)=x^3+5x^2-17x-21

Answer 1

I know I have to find the x value first and then plug it into g(x) but I'm not sure how to get the x value.

Answer 2

you doing calculus or plotting graphs ?!?!

Answer 3

precalc

Answer 4

assume that means you are not allowed to differentiate

but your thing factorises....

\((x + 1) (x - 3) (x + 7)\)


and looks like this....




so that might even help

Answer 5

But then how do I take that information and use it to find the local maximum value

Answer 6

you know what....right now i haven't a clue how to help you without using calculus :(

Answer 7

Use the calculus I'm willing to try it

Answer 8

here is an interesting way

Answer 9

shift your graph up until you get a double root, then you have found your minimum

Answer 10

you want your root at -1 and 3 to coalesce

Answer 11

Hang on @sharkfinsoup, smart dude at work !!

Answer 12

ok take your time @IrishBoy123

Answer 13

What do I do after I shift the graph? @inkyvoyd

Answer 14

g(x)=x^3+5x^2-17x-21
g(x)+a=x^3+5x^2-17x-21+a=(x+7)(x-b)^2
so \(x^3+5x^2-17x-21+a=(x+7)(x-b)^2\)
\(x^3+5x^2-17x-21+a=7 b^2 - 14 b x + b^2 x + 7 x^2 - 2 b x^2 + x^3\)
so let's try to solve for a and b...

Answer 15

oops made a mistake... we don't know the root will stay at x=-7 unfortunately

Answer 16

so actually,
\(g(x)+a=x^3+5x^2-17x-21+a=(x+b)(x-c)^2\)

Answer 17

so
\(x^3+5x^2-17x-21+a=b c^2 - 2 b c x + b x^2 + c^2 x - 2 c x^2 + x^3\)
Now simplify:
\(5x^2-17x-21+a=b c^2 - 2 b c x + b x^2 + c^2 x - 2 c x^2\)

Answer 18

Group terms:
\(0=(b-2c-5)x^2+(17-2bc+c^2)x+(21-a+bc^2)\)

Answer 19

oh gosh darnit all my progress goes toward finding the local minimum LOL...
well that is a good thing in the end since you can use my method to figure out the local max on your own

Answer 20

So LHS=0=RHS=0 so all coefficients must be 0... We need to solve a yucky system of 3 equations in 3 unknowns with quadratics, but it's doable

Answer 21

0=b-2c-5
0=17-2bc+c^2
0=21-a+bc^2

Answer 22

if you do a ton of algebra you will get
http://prntscr.com/di06yr

Answer 23

so our minima must be at x=a = \(-\frac{16}{27}(28+19\sqrt{19})\)
or something of the sort...

Answer 24

that's not a nice result.

Answer 25

and it doesn't agree with the plot.

Answer 26

might be 28-19sqrt(19)

Answer 27

\(-\frac{16}{27}(28-19\sqrt{19})\)
there we go, your local minima

Answer 28

FYI @UnkleRhaukus I did the calculus and it is not a nice result. it IS my result though so I'm happy.

Answer 29

but fyi sharkfin your numbers are so nasty that I wouldn't want to do this problem if calculus WAS allowed.

Answer 30

you had \(x = −\frac{16}{27}(28+19\sqrt{19}) \),
i think you want \(y = \frac{16}{27}(28+19\sqrt{19}) \)

Answer 31

Alright well for anyone curious the correct answer was 65.67