On a coordinate plane, a curved line with a minimum value of (negative 2.5, negative 12) and a maximum value of (0, negative 3)
2 answers:
Answer:
f(x) > 0 over the interval
Step-by-step explanation:
If f(x) is a continuous function, and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.
Then the function was above the x axis (larger than zero) from , until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.
There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from to -3, since between the values -4 and -3 the function adopts negative values.
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Looks like .
- If
, then
- critical point at (0, 0).
- If
, then
- two critical points at
and
The latter two critical points occur outside of since
so we ignore those points.
The Hessian matrix for this function is
The value of its determinant at (0, 0) is , which means a minimum occurs at the point, and we have
.
Now consider each boundary:
- If
, then
which has 3 extreme values over the interval of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).
- If
, then
and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).
- If
which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).
- If
, then
which has 3 extreme values, but the previous cases already include them.
Hence has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).