Function Analysis

Question

I tried to differentiate the function and set the value at =0 in order to find x in terms of a and b. Then, I also tried to plug the value of x into in and set the value of y into 1 and 4. At the end, I was struggling since it became more tedious. So anyone can suggest a nicer way to solve this? Thanks in advance

Answer 1

I found a solution which is a little bit tedious but less than some of the others.

The equation $f(x)=1$ is equivalent to: \begin{align*} ax^2 + 4x + b &= x^2 + 2 \\\iff (a-1)x^2 + 4x + (b-2) &= 0 \end{align*} This is a quadratic equation in $x$. If $f$ has a minimum at $x$ too, then we can assume the equation has one double solution. Therefore the discriminant of the quadratic polynomial must be zero: \begin{align*} 0 &= 4^2 - 4(a-1)(b-2) = 4(2a+b + 2 - ab) \end{align*} In a similar way, the equation $f(x) = 4$ gives \begin{align*} ax^2 + 4x + b &= 4(x^2 + 2) \\\iff (a-4)x^2 + 4x + (b-8) &= 0 \end{align*} Since $4$ is the maximum value, we expect a double solution, so the discriminant is zero as well: \begin{align*} 0 &= 4^2 - 4(a-4)(b-8) = 4(8a+4b - 28 - ab) \end{align*} Putting these together, we have two equations in the two unknowns $a$ and $b$: \begin{align*} 2a+b - ab &= -2 \tag{1} \\ 8a + 4b - ab &=28 \tag{2} \end{align*} Subtracting ($1$) from ($2$) gives $$ 6a+3b = 30 \implies 2a+b = 10 \tag{3} $$ Subtracting $4$ times ($1$) from ($2$) gives $$ 3ab = 36 \implies ab = 12 \tag{4} $$ We can substitute ($3$) into ($4$) to get \begin{align*} a(10-2a) = 12 \implies a^2 - 5a + 6 = 0 \end{align*} The solutions are $a = 2$ (yielding $b=6$) and $a=3$ (yielding $b=4$).

Answer 2

HINTS

The only way I could get through the tedium was using wolfram. I solved the problem as follows.

You have two equations

$\quad f(x) = 4 \text{ and } f'(x) = 0$

in three variables, $a$, $b$ and $x$.

So attempt to eliminate $a$ or $b$ and when you do you'll find that $b$ can be expressed in terms of $a$,

$\tag 1 b = \frac{4 (2 a - 7)}{a - 4}$

Use Wolfram Step 1 which solves for $b$.

After a copy-paste, you have two equations,

$\quad f(x) = 1 \text{ and } f'(x) = 0$

in two variables, $a$ and $x$.

Here you'll get to choose between two $a,x$ solutions and only one will work.

Use Wolfram Step 2 which solves for $a$ and $x$.

Answer 3

$$\text{max}\big\{f(x) \big\} = \frac{2a+b+\sqrt{32 + (b-2a)^2}}{4} \qquad \text{min}\big\{f(x) \big\} = \frac{2a+b-\sqrt{32 + (b-2a)^2}}{4}$$

So either $a = 3$ and $b = 4$ or $a = 2$ and $b = 6$.