Question

f(x)=x^3-3x^2-9x+4 find the intervals on which f is increasing or decreasing b. find the local maximum and minimum values of f. c. find the intervals of concavity and inflection points

Answer 1

Answer:

Please read the complete answer below!

Step-by-step explanation:

You have the following function:

$f(x)=x^3-3x^2-9x+4$   (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:

$\frac{df}{dx}=3x^2-6x-9$    (2)

Next, you equal the derivative to zero, and then you find the roots of the polynomial by using the quadratic formula:

$3x^2-6x-9=0\\\\x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(3)(-9)}}{2(3)}\\\\x_{1,2}=\frac{6\pm12}{6}\\\\x_1=-1\\\\x_2=3$

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

$\frac{df(-1.01)}{dx}=3(-1.01)^2-6(-1.01)-9=0.12$

Then, in the interval (-∞,-1), the function is increasing

for x = -0.99

$\frac{df(-0.99)}{dx}=3(-0.99)^2-6(-0.99)-9=-0.11$

In the interval (-1,3) the function is decreasing

for x = 3.01

$\frac{df(3.01)}{dx}=3(3.01)^2-6(3.01)-9=0.12$

In the interval (3,+∞) the function is increasing

b) To find the local minimum and maximum you use the second derivative of the function:

$\frac{d^2f}{dx^2}=6x-6$     (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

$6(-1)-6=-12$

x=-1 is a local maximum

for x2 = 3

$6(3)-6=12>0$

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:

$6x-6=0\\\\x=1$

For x = 1 you have an inflection point