Question

The function f is defined by `f(x)=x^4-4x^2+x+1` for `-5<=x<=5`. What is the interval in which the minimum of value of f occur?

Answer 1

Purely a graphical approximation; Minimum f = -4.63, nearly. This is improved to 8-sd,`-4.4441919`, using an iterative numerical method.

Explanation:

f-graph reveals approximations to all the four [zeros].

(https://socratic.org/precalculus/polynomial-functions-of-higher-

degree/zeros) of f in (-3, 2)

and the minimum occurs, in between negative zeros,

graph{x^4-4x^2+x+1 [-10, 10, -5, 5]}

`f'=4x^3-8x+1`.

The f'-graph reveals its zero near -1.55, for the turning point.

The third f-graph for tangency, with the tangent line, reveals

horizontal tangent at `f = -4.63`, nearly.

graph{4x^3-8x+1x^2 [-1.55, 0, -10, 10]}

graph{(x^4-4x^2+x+1-y)(y+4.63)=0 [-2.06, 0, -10, 10]}

This zero of f' is improved to 8-sd,`-1.4729875`, using an iterative

numerical method.

So, the required minimum is

`f(-1.4729875)`

`=-4.4441919`

Note: There are limitations to accuracy, in graphical approximations.