Please help!!
Okay so my teacher has been away and hasent explained anything to us and we have to find
the
intercepts,
critical points,
intervals of increase and decrease,
points of inflection,
intervals of concavity,
local maximum or minimum points
for each of the following functions
Can you please help me?
1.g(x) = x^3 − 6x^2 + 9x
2.p(x) = x^4 − 4x^3
So far i have the first derivative of each
3x^2-12x+9
4x^3-12x^2
for intercepts, set x=0 and evaluate y for the y-intercept
Set y=0 and solve for x to get the x-intercepts. Note that
g(x) = x(x-3)^2
p(x) = x^3(x-4)
Max/min occurs when 1st derivative = 0. No problems there, right?
But is it a max or a min?
If the curve is concave up, it's obviously a minimum.
So, check the 2nd derivatives. Where
f" < 0 the curve is concave down
f" > 0 the curve is concave up
f" = 0 concavity changes: inflection point. No max or min.
well first of all it asked for the intercepts
where does the first one hit the x axis?
x (x^2 -6 x+ 9) = 0
x (x-3)(x-3) = 0
so
x axis intercepts at x = 0 and x = 3 (twice which means that is a grazing or min or max of g
Where does the first one hit the g axis?
when x = 0, y = 0 so at the origin
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critical points (we already know about at x = 3)
0 = 3 x^2 - 12 x + 9
0 = 3 (x^2 -4 x + 3)
(x-3)(x-1) = 0
so critical points at x = 1 and x = 3
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left of x = 1, function is increasing because x^3 is -oo at x =-oo
at x = 1, g is 4
between x = 1 and x = 3, function is decreasing from 4 to 0
right of x = 3, function is increasing because for x=+oo, g = +oo
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SKETCH THE GRAPH NOW, max of 4 at x = 1 and min of 0 at x = 3
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Now you do the second function :)
p = x^3 (x-4)
zero at x = 0 and at x = 4
p' = 4 x^3 - 12 x^2
horizontal at
0 = 4 (x^3 - 3 x^2 ) = x^2 (x-3)
so critical at x = 0 and x = 3 etc