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.

no he doesnt because "-16' is negatiive

step-by-step explanation: