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December 21, 2014

December 21, 2014

Posted by **Derek** on Tuesday, July 29, 2008 at 8:45pm.

ƒ(x) = −x^2 + 4x − 3

g(x) = x^3 − 6x^2 + 9x

h(x) = x^4 − 6x^2

p(x) = x^4 − 4x^3

- URGENT! Last calculus question! -
**Damon**, Tuesday, July 29, 2008 at 9:16pmI will do part of one but really think you should try it yourself first and then ask for help with specific questions.

In general look for what it says to look for.

For example the second one

g(x) = x^3 - 6 x^2 + 9x

first of all what happens to this for large negative x and for large positive x?

for large x magnitude it will look like x^3 so it will go off to negative infinity for large negative x and off to positive infinity for large positive x, sloping ever more steeply (concave)

Now where will it be zero?

set it equal to zero

x^3 - 6 x^2 + 9x = 0 = x(x^2-6x+9)=x(x-3)(x-3)

so zero if x = 0 and double zero (just touches the x axis, does not go through) at x = 3

Now we have a pretty good idea what it looks like already.

It snakes up from -oo on the left, goes through (0,0) headed up to the right, then dips down to the x axis, just touching, before heading up to the right.

Now where is it horizontal between x = 0 and x = 3 ?

take derivative and set to zero

dg/dx = 3 x^2 -12 x + 9

set to zero and factor out 3

0 = x^2 -4x + 3

0 = (x-1)(x-3)

so horizontal at x = 1 (and as we knew at x=3)

What is g at that extreme at x = 1?

g = 1^3 - 6*1^2 + 9*1

=1 - 6 + 9

= 4

That should about do it, you can sketch it pretty neatly

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