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.