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?

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Answer 1 · 2016-12-31T16:38:28

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

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'=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.