# Maximizing $y=|4x^3+ax^2+bx+c|$ in $[-1,1]$

• MHB
• anemone
In summary, the maximum value of y in the given interval is determined by finding the critical points of the function and evaluating the function at those points. The coefficients a, b, and c in the given function affect the shape and position of the graph of the function, which can change the maximum value of y. The maximum value of y can be negative in the given interval, but there can only be one maximum value due to the continuity of the function. Calculus can be used to find the maximum value of y by finding the critical points and using the second derivative test.
anemone
Gold Member
MHB
POTW Director
Let $a,\,b$ and $c$ be real numbers and let $K$ be the maximum of the function $y=|4x^3+ax^2+bx+c|$ in the interval $[-1,1]$. Show that $K\ge 1$. For which $a,\,b$ and $c$ is the equality occurs?

Let $p(x) = -ax^2 - bx - c$, so $p(x)$ is a quadratic (or lower degree) polynomial, and we want to find the smallest possible value of $K = \max\{|4x^3 - p(x)|:-1\leqslant x\leqslant 1\}.$

Notice that if $|4x^3 - p(x)| \leqslant K$ for $-1\leqslant x\leqslant 1$, then $|4(-x)^3 - p(-x)| = |4x^3 + p(-x)| \leqslant K$ for $-1\leqslant x\leqslant 1$. Therefore $|4x^3 - q(x)| \leqslant K$ for $-1\leqslant x\leqslant 1$, where $q(x) = \frac12(p(x) - p(-x))$. But $q(x)$ is an odd function. So to find the function that minimises $K$ we need only look at odd functions, and the only odd polynomials of degree at most $2$ are multiples of $x$.

Therefore in the polynomial $p(x) = -ax^2 - bx - c$ we should take $a=c=0$, and it is easy to see that the optimum value of $b$ is $-3$. Then (as in the diagram below) the graph of $y= 4x^3$ lies between the lines $y = 3x\pm1$ in the interval $[-1,1]$, and $1 = \max\{|4x^3 - 3x|:-1\leqslant x\leqslant 1\}$ is the smallest possible value for $K$.

[TIKZ][xscale=4,yscale=5]
\draw [step=0.25cm, help lines] (-1.1,-1.1) grid (1.1,1.1) ;
\draw (-1.1,0) -- (1.1,0) ;
\draw (0,-1.1) -- (0,1.1) ;
\draw[very thick, domain=-1.1:1.1] plot (\x,{\x^3});
\draw [thick] (-1.1, -0.825) -- (1.1,0.825) ;
\draw [dashed,thin] (-1.1, -1.075) -- (1.1,0.575) ;
\draw [dashed,thin] (-1.1, -0.575) -- (1.1,1.075) ;
\foreach \x in {-1,0.5,1} \draw (\x,-0.07) node [fill=black!8] {$\x$} ;
\foreach \y in {-4,-3,...,4} \draw (-0.13,\y/4) node [fill=black!8] {$\y$} ;[/TIKZ]

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## 1. What is the maximum value of y in the given range?

The maximum value of y in the given range can be found by finding the critical points of the function and evaluating them within the range. The critical points can be found by taking the derivative of the function and setting it equal to 0. Once the critical points are found, plug them back into the original function to determine the maximum value of y.

## 2. How do the coefficients of the function affect the maximum value of y?

The coefficients a, b, and c in the function have a direct impact on the shape and position of the graph. Changing these coefficients can shift the graph horizontally or vertically, making it easier or harder to find the maximum value of y. The larger the coefficients, the steeper the graph will be, making it more challenging to find the maximum value.

## 3. Can the maximum value of y be negative?

Yes, the maximum value of y can be negative. The absolute value function ensures that the output is always positive, but it does not restrict the maximum value from being negative. In some cases, the maximum value of y can be a negative number, depending on the coefficients of the function.

## 4. Can the maximum value of y occur at a point other than the critical points?

Yes, the maximum value of y can occur at a point other than the critical points. This can happen if the function is not continuous within the given range or if there are multiple critical points with the same value. In these cases, it is essential to check all possible points within the range to determine the maximum value of y.

## 5. Is there a shortcut or formula for finding the maximum value of y in this type of function?

No, there is no specific formula or shortcut for finding the maximum value of y in this type of function. The process involves taking the derivative, finding the critical points, and evaluating them within the given range. However, some techniques, such as using the first or second derivative test, can make the process more efficient and accurate.

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