Question

What is the maximum number of possible extreme values for the function,

Answer 1

Answer:

(0.3, -18.45).

Step-by-step explanation:

We need to recur to the extreme value theorem, which states: "If a function is continuous on a closed interval, then that function has a maximum and a minimum inside that interval".

Basically, as the theorem states, if a dunction is continuous, then it has maxium or minium.

In this case, we have a quadratic function, which is a parabola. An important characteristic of parabolas is that they have a maximum or a minium, but they don't have both. When the quadratic term of the fuction is positive, then it has a minium at its vertex. When the quadratic term of the function is negative, then it has a maximum at its vertex.

So, the given function is $f(x)=x^{2} +4x^{2} -3x-18=5x^{2} -3x-18$, where the quadratic term is positive, so the functions has a minimum at $V(h,k)$, where $h=-\frac{b}{2a}$ and $k=f(h)$, let's find that point

$h=-\frac{-3}{2(5)} =\frac{3}{10} =0.3$

$k=f(0.3)=5(0.3)^{2} -3(0.3)-18=0.45-0.9-18=-18.45$

Therefore, the minium of the function is at (0.3, -18.45).