Question

Question #5a417

Answer 1

Local minimum `f(2) = d - 44`.

Explanation:

`f'(x) = 6x^2+6x+c`
`f''(x) = 12x + 6`

f(2) is a local minimum, if f'(2) = 0 and f''(2) > 0.

`f''(2) = 30 > 0 and f'(2) = 0 to c= -36`.

Now, the local minimum `f(2) = -44+ d`.

This following is my model for f(x;a;E), a function with local extremum E, at x = a.

Here, the parameters a and E are at your choice.'

#f(x:a:E) = E +-(x-a)g(x), where g(a) = 0 and g'(a)> or < 0.
So, the design of g(x) is subject to the stated conditions. Yet, it is at your choice.

f(a) = E, the local minimum for + sign prefixing (x-a) and this sign that corresponds to g'(a) < 0.

E is the local maximum, in the other case.

Illustrative example: a = 1 and E = 5. Choose g(x) = x(x-1).

f(x;1;5) for local minimum 5 at x - 1 is
#5 + x(x-1)^2

Here g(x) = +x(x-1). g(1) = 0 and g'(1) = 1 > 0.
The local minimum at x = 1 is f(1;1;5) = 5.
For 5 to be the local maximum, g(x) =`- x(x-1)`. ..