# 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?

Dec 31, 2016

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].

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]}

$f ' = 4 {x}^{3} - 8 x + 1$.

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

$f \left(- 1.4729875\right)$

$= - 4.4441919$

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