A Question about a function's parameters given it's minimum value for a certain domain

Question

So the question goes like this: Find all the values of a , for which f(x)=4x^2 -12ax + 9a^2 +3a -2 function's minimum value in [0: 2] domain equals 4.

What am I supposed to do with the given information? Even If I found X0 or Y0, I can't seem connect it with [0: 2]. I have never done this type of problem before. Any help would be appreciated, Thanks.

Answer 1

Clearly, we are interested in finding the maxima of a function. There is a standard procedure for this, where we must check three things:

1. Discontinuities: As a polynomial, this function does not have discontinuities in its domain.
2. End point of the domain: We must calculate $f(0)$ and $f(2)$, which turn out to be $f(0) = 9a^2 + 3a -2 = C_1$; and $f(2) =9a^2 -21 a +14 = C_2$
3. Places where the derivative is zero: The derivative of the function is $\frac{df(x)}{dx} = 8x - 12a$, therefore the derivative has a root at $x= \frac{3}{2}a$; the value of the function at this point is $f(x=\frac{3}{2}a) = 3a-2= C_3$

Depending on the value of $a$, the global minimum of the function can be any of these three candidates $C_1, C_2$ or $C_3$. Now we only need to find all the values of $a$ so that either $C_1 = 4$, $C_2 = 4$ or $C_3 = 4$:

• $C_1= 4 \Leftrightarrow 9a^2 + 3a -2 = 4 \Leftrightarrow a=-1 ~\text{or}~a= 2/3$
• $C_2 = 4 \Leftrightarrow 9a^2 -21 a +14 = 4 \Leftrightarrow a=2/3 ~\text{or}~a=5/3$
• $C_3 = 4 \Leftrightarrow 3a -2 = 4 \Leftrightarrow a = 2$

Now you still need to compare the three values at each of the four points, to weed out the ones that actually result in a maximum rather than a minimum. Can you do the final point?