Question 1079044
A stationary point is where the first derivative is defined and equal to zero
:
f'(x) = 3x^2 + p
:
we know x = -2 at the stationary point (-2,3)
:
1)  3(-2)^2 + p = 0
p = -12
:
we know f(-2) = 3 since (-2,3) is a point on the graph of f(x)
:
2) 3 = (-2)^3 -12(-2) + q
3 = -8 +24 + q
q = -13
:
f(x) = x^3 -12x -13
:
**************************
a.  p = -12, q = -13
**************************
:
f'(x) = 3x^2 -12
:
3x^2 -12 = 0
:
3x^2 = 12
:
x^2 = 4
:
x = 2 and x = -2
:
***************************
b. for x = 2
f(2) = 2^3 -12(2) -13 = -29
stationary point at (2,-29)
***************************
:
f'(x) = 3x^2 -12
:
An inflection point is a point on a curve where the curve changes sign.
:
we take the second derivative, which is
:
f''(x) = 6x
:
our candidate x value is x = 0
:
Note that if x is negative then 6x is negative and if x is positive then 6x is positive, therefore the point with x = 0 is an inflection point
:
f(0) =  0^3 -12(0) -13 = -13
:
**************************************
c.  the point of inflection is (0,-13)
**************************************
:
the graph of f(x) = x^3 -12x -13 looks like
:
{{{ graph( 300, 200, -5, 5, -50, 40, x^3 -12x -13) }}}
:
we want values of k, such that f(x) crosses the x axis 3 times
:
***************************************************************
d.  looking at the graph, this is where f(x) = 0, that is, k = 0
***************************************************************
: