Answer:
Please read the complete answer below!
Step-by-step explanation:
You have the following function:
[tex]f(x)=x^3-3x^2-9x+4[/tex] (1)
a) To find the interval on which f is increasing or decreasing, you first calculate the critical points of f(x).
You calculate the derivative f(x) respect to x:
[tex]\frac{df}{dx}=3x^2-6x-9[/tex] (2)
Next, you equal the derivative to zero, and then you find the roots of the polynomial by using the quadratic formula:
[tex]3x^2-6x-9=0\\\\x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(3)(-9)}}{2(3)}\\\\x_{1,2}=\frac{6\pm12}{6}\\\\x_1=-1\\\\x_2=3[/tex]
Then, the critical points are x=-1 and x=3
Next, you calculate df/dx for a values of x to the left and to the right of the critical points x1 and x2. If df/dx < 0 the function is decreasing, if df/dx > 0 the function is increasing.
for x = -1.01
[tex]\frac{df(-1.01)}{dx}=3(-1.01)^2-6(-1.01)-9=0.12[/tex]
Then, in the interval (-∞,-1), the function is increasing
for x = -0.99
[tex]\frac{df(-0.99)}{dx}=3(-0.99)^2-6(-0.99)-9=-0.11[/tex]
In the interval (-1,3) the function is decreasing
for x = 3.01
[tex]\frac{df(3.01)}{dx}=3(3.01)^2-6(3.01)-9=0.12[/tex]
In the interval (3,+∞) the function is increasing
b) To find the local minimum and maximum you use the second derivative of the function:
[tex]\frac{d^2f}{dx^2}=6x-6[/tex] (3)
you evaluate the second derivative for the critical points x1 and x2, if the second derivative is positive, you have a local minimum. If the second derivative is negative, you have a local maximum:
for x1 = -1
[tex]6(-1)-6=-12<0[/tex]
x=-1 is a local maximum
for x2 = 3
[tex]6(3)-6=12>0[/tex]
x=3 is a local minimum
c) upward concavity: (-1,3)
downward concavity: (-∞,-1)U(3,+∞)
The inflection points are calculated with the second derivative equal to zero:
[tex]6x-6=0\\\\x=1[/tex]
For x = 1 you have an inflection point
