Find the relative minimum of this equation

Step-by-step explanation:

Given: f(x) = y = 3*x^3+14*x^2-11*x-46 .....................(1)

To find: relative minimum.

To find the relative minima of a polynomial f(x), we examine the values of its derivative, f'(x).  

A necessary condition for a minimum (or maximum) to exist is if at least one root exists for f'(x) = 0.

In this case,

f'(x) = 3*3x^2 + 2*14x - 11 = 9x^2 + 28x - 11  ...................(2)

Equate the derivative, or equation (2) to zero and solve for x using the standard quadratic formula, we get

x1 = -(sqrt(295)+14)/9,   or   ........................(3a)

x2 = (sqrt(295)-14)/9 ....................................(3b)

To check if the solutions are maximum or minimum, we calculate the second derivative, f"(x).

f"(x) >0 => f(x) is a minimum

f"(x) =0 => f(x) is an inflexion point

f"(x) <0 => f(x) is a maximum.

from (2), we derive to get

f"(x) = 18x+28

We check each of x1 and x2

f"(x1) = f"( -(sqrt(295)+14)/9 ) = -34.35   => x1 is a maximum.

f"(x2) = f"((sqrt(295)-14)/9) = 34.35       => x2 is a minimum.

Hence we deduce that x2 = (sqrt(295)-14)/9 = 0.35284 is a minimum.

See graph attached.

step-by-step explanation:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since     2010 0     1     2     3       4       5

profits made in 1000's   39       84     111     120   111     84

we find that profit gradually increases form 2010 to 2013 upto 120,   then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

read more on -