Part a)
The given polynomial function is
[tex]f(x) = {x}^{3} - 9 {x}^{2} - 21x + 8[/tex]
The first derivative is :
[tex]f'(x) = {x}^{2} - 18x - 21[/tex]
This is the slope function.
On an increasing interval, the slope is greater than 0.
[tex]6 {x}^{2} - 18x - 21 \: > \: 0[/tex]
This gives;
[tex]x \: < \: - 1 \: or \: x \: > \: 7[/tex]
Similarly the function is decreasing when
[tex]6 {x}^{2} - 18x - 21 \: > \: 0[/tex]
This gives:
[tex] -1 \: < \: x \: < \:7[/tex]
Part b)
At fixed points,
[tex]f'(x) = 0[/tex]
This implies that:
[tex]3 {x}^{2} - 18x - 21 = 0[/tex]
[tex](x + 1)(x - 7) = 0[/tex]
[tex]x = - 1 \: or \: x = 7[/tex]
We apply the second derivative test.
[tex]f''(x) = 6x - 18[/tex]
[tex]f''( - 1) = 6( - 1) - 18 = - 24 \: < \: 0[/tex]
Hence (-1,19) is a local maximum.
[tex]f''(7) = 6(7) - 18 = 24 \: > \: 0[/tex]
This means (7,-237) is a local minimum.
Part c)
At the point of inflection, the second derivative is zero.
This implies that;
[tex]f''(x) = 6x - 18 = 0[/tex]
We solve for x to get:
[tex]x = 3[/tex]
[tex]f(3) = {3}^{3} - 9( {3}^{2} ) - 21(3) + 8 = - 109[/tex]
The point of inflection is (3,-109)
Part d)
The given polynomial function is concave up on the interval where f''(x)>0.
This means the interval of concavity is on 6x-18>0.
6x>18
x>3
The function is concave up on (3,∞).
Part e)
The given function is concave down when f''(x)<0.
This implies that;
6x-18<0.
6x<18
x<3
The interval on which the function is concave down is (-∞,3)
