Is this always minimum?

Calculus Level 5

Consider a function $f\colon\Bbb R\to\Bbb R$ defined by $f(x)=x^{3}+3x^{2}+4x+b\sin(x)+c\cos(x)$

Given that $f$ is one-to-one , find the maximum value of $b^{2}+c^{2}$ .

The answer is 607.6.

1 solution

Calvin Lin Staff
Jan 28, 2015

[This is not a rigorous solution.]

Since the function is one-to-one, we know that the derivative must be non-negative or non-positive. We calculate that

$f'(x) = 3x^2 + 6x + 4 + b \cos x - c \sin x$ .

For fixed value of $b$ and $c$ , for large enough value of $x$ , we have $3x^2 + 6x + 4 > |b| + |c| > - b \cos x + c \sin x$ . This tells us that we want $f'(x) \geq 0$ .

Now, let $b \cos x - c \sin x = A \cos ( x + \theta)$ , where $A = \sqrt{ b^2 + c^2 }$ and $\theta = \tan^{-1} \frac{c}{b}$ . Thus, we want to maximize $A$ subject to $3x^2 + 6x + 4 \geq - A \cos ( x + \theta)$ , for some $\theta$ .

(This is where the solution stops being rigorous.) To determine $\theta$ , it makes sense that the minimum of $- \cos ( x + \theta )$ (red line) coincides with the minimum of $3x^2 + 6x + 4$ (yellow line). After that, we scale the graph so that $- A \cos (x + \theta )$ (blue line) is tangential to the parabola, so that we cannot increase $A$ further.

Since the minimum occurs at $x = -1$ , we thus want $x + \theta = 0$ , and so $\theta = 1$ .

We now want to maximize $A$ , and this is satisfied when the yellow and blue lines are equal and tangential at 2 points which are distance at most $2 \pi$ apart. This leads to the system of equations

$\begin{cases} 3x^2 + 6 x + 4 = - A \cos (x + 1) & \text{ from equality} \\ 6x + 6 = A \sin ( x + 1 ) & \text{ from tangency}\\ \end{cases}$

Make the substitution $y = x + 1$ , and the system becomes

$\begin{cases} 3y^2 + 1 = - A \cos y \\ 6y = A \sin y \\ \end{cases}$

At the solution points, we must have $\tan y = - \frac{ 6y}{ 3y^2 + 1 }$ . There is (likely) no way to solve this analytically, so we resort to Wolfram and obtain $y = \pm 2.49$ . Substituting this in, we get $A = \frac{ 6 \times 2.49 } { \sin 2.49 } \approx 24.65$ .

Hence, the maximum of $b^2 + c^2 = A^2 = 24.65^2 \approx 607.6$ .

Calvin, this is the same answer that I got. I've decided to check this out after reviewing "What's the minima?" note posted elsewhere.

Michael Mendrin - 6 years, 4 months ago

This leads to the system of equations...

Can you explain this part more in detail?

Daniel Liu - 6 years, 4 months ago

The points lie on both graphs, hence $3x^2 + 6x+4 = - A \cos (x+1)$ .
The points have the same tangents, hence $6x + 6 = A \sin (x+1)$ .

Calvin Lin Staff - 6 years, 4 months ago

$\text {I think The answer should be 1 , Look at my solution } \\ \text{First Let's take the Derivative of the Function } \\ f^{'}(x)=3x^2+6x+4+b\cos{x}-c\sin{x} \\ \text {Now, for the function to be one-one the Derivative should be always positive } \\ \text{Because coefficient of } x^2 \text{ is positive so }f^{'}(x)>0\\ \text{To make } b\cos{x}-c\sin{x} \text{ Always Positive,Add } \sqrt{b^2+c^2} \text{ to it.} \\ \text{so , the rest of the function , that is } 3x^2+6x+4-\sqrt{b^2+c^2} \ge{0} \\ \text{Discriminant(D) } \le{0} \\ \implies 36-12(4-\sqrt{b^2+c^2})\le{0} \\ \implies 3-4+\sqrt{b^2+c^2}\le{0} \implies \text {Max. Value of } \color{#3D99F6}{\boxed{b^2+c^2=1}}$

Sabhrant Sachan - 5 years, 1 month ago

I agree that the derivative should have the same sign, and we must have $f'(x) \geq 0$ .

However, the rest of the solution doesn't apply. You have replaced $b\cosx - c \sin x$ with a much smaller value of $- \sqrt{b^2 + c^2 }$ . As such, you have found a sufficient condition such that $f'(x) \geq 0$ . However, that isn't a necessary condition, nor does it search out the maximum value of $b^2 + c^2$ .

Calvin Lin Staff - 5 years, 1 month ago

To make the function easier to solve i have added and subtracted $\sqrt{b^2+c^2}$ so that $b\cos{x}-c\sin{x}$ is always positive for all x belonging to real, because $-\sqrt{b^2+c^2}\le b\cos{x}-c\sin{x}\le\sqrt{b^2+c^2} \\ \implies 0\le \sqrt{b^2+c^2}+b\cos{x}-c\sin{x}\le2\sqrt{b^2+c^2}$

After these changes we can ignore $\sqrt{b^2+c^2}+b\cos{x}-c\sin{x}$ and concentrate on making the rest of the function positive, that is $3x^2+6x+4-\sqrt{b^2+c^2} \ge{0}$ and for that the discriminant must be less than 0

Sabhrant Sachan - 5 years, 1 month ago

@Sabhrant Sachan – Yes, I understand what you are thinking.

What you are saying is that "Suppose I want to make $f(x) + g(x) \geq 0$ . Then, because I know that $g(x) \geq h(x)$ , what I will do is find when $f(x) + h(x) \geq 0$ . Under such a condition, it will be true that $f(x) + g(x) \geq f(x) + h(x) \geq 0$ , and thus this is a sufficient condition. However, since it's possible for $f(x) + g(x) \geq 0 > f(x) + h(x)$ , thus the condition isn't necessary.

Put another way, if I asked you to "Find the minimum value of $|x|$ such that $x^2 - 1 \geq 0$ ", can you say that "Since $-1 > -100$ , hence I will solve $x^2 - 100 \geq 0$ and then conclude that the answer is only $|x| \geq 10$ thus the minimum is $|x|$ ?". No, all that we have is found a sufficient condition, which might not include all possible cases. As such, we cannot conclude that we have indeed found the extrema.

Calvin Lin Staff - 5 years, 1 month ago

@Calvin Lin – Thanks sir :) , i now understand my mistake .

Sabhrant Sachan - 5 years, 1 month ago

Is this always minimum?

The answer is 607.6.

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