Find Absolute Minimum From Graph Attributes
Consider a function f with the following properties:
1.) f is continuous and differentiable on R.
2.) $f'(x) = 0 \implies x = -2$ or $ x = 1$
3.) The following points are given:
$(-3,5);(-2,0);(-1,2);(1,4);(2,2);(3,1.5)$
From this I made the following assumptions , f is a polynomial function ( on account of 1 ) ; At x = -2 and x = 1 the gradient is equal to zero , meaning that the function reaches a maximum/minimum at these locations.
Looking at the given points I assumed that the function is decreasing from -3 to -2 ; Increasing from -1 to 1 ; Decreasing from 2 to 3. In these points 1.5 is also the smallest y value in the range [-1,3].Could it then be assumed that 1.5 is the absolute minimum?
A video tutorial mentioned that you evaluate the critical and end points thus and simply find the smallest y value:
$f(-1)=1 ; f(-2)=0 ; f(1)=4 ; f(3)=1.5$
Where once again f(3) has the smallest y value in the range [-1,3].
$\endgroup$1 Answer
$\begingroup$First, just because the first derivative is zero does not mean you have a maximum or minimum. For example consider $x^3$, that is
$$\frac{d}{dx}x^3 = 3x^2$$
which has a zero (ie a critical point) at $x = 0$, but if you look at a graph of $x^3$ there clearly isn't a maximum or a minimum at $x=0$
In your case, your given that the first derivative is zero at $x = -2$ and the points of $f$ you gave show that $f$ attains a minimum there. Hence by the extreme value theorem there is a local minimum at this point.
Restricting our attention to the domain of the points given clearly indicates that the absolute minimum occurs at $x = -2$ (i.e. $0 < 1.5$)
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