Animated Solution for Mathematics - Differentiation: Let $P(x)=a_0+a_1{x}^2+a_2{x}^4+\cdots +a_n{x}^{2n}$ be a polynomial in a real variable $x$ with $0<a_0<a_1<a_2<\cdots <a_n$. The function $P(x)$ has

Unlocking the Polynomial

Welcome, fellow explorer of mathematics. Today, we are not just solving a polynomial problem; we are peeling back the layers of a function to reveal its geometric soul.

When you first look at $P(x)=a_0+a_1{x}^2+a_2{x}^4+\cdots +a_n{x}^{2n}$, it might seem like a daunting, long-winded expression. But I want you to pause and look past the complexity. Do you see the pattern? Every single power of $x$ is even. This is not a coincidence; it is the key to the entire kingdom.

The Power of Evenness

In the world of functions, an even function is a beautiful thing. It possesses a mirror-like quality. Because $P(x)=P(-x)$, the graph is perfectly symmetric about the $y$-axis.

Imagine standing at the origin and looking left and right; the landscape is identical. This symmetry is our first clue. It tells us that if there is a minimum or maximum, it is likely tied to the origin, $x=0$. But we cannot rely on intuition alone; we must prove it with the rigor of calculus.

The Derivative Hunt

To find the extrema—the peaks and valleys of our function—we need to find where the slope is zero. We invoke the first derivative, $P^{\prime }(x)$. Differentiating term by term, we get:

I know what you are thinking: "That looks like a mess." But look closer! Every single term shares a common factor of $x$. Let us pull that $x$ out into the light:

The Bracketed Mystery

Let us define the expression inside the bracket as $B(x)$. This is where the magic happens. We are given that $0<a_0<a_1<a_2<\cdots <a_n$. All coefficients are strictly positive.

Furthermore, every power of $x$ inside $B(x)$ is even. We know that for any real number $x$, an even power is always non-negative. When you multiply positive coefficients by non-negative powers and add them up, the result is always positive.

Even at $x=0$, the bracket evaluates to $2a_1$, which is strictly greater than zero. Therefore, $B(x)>0$ for all real $x$. This is a profound realization. Our derivative $P^{\prime }(x)=x\cdot B(x)$ can only be zero if $x=0$. There are no other critical points. The function has only one stationary point in its entire domain.

The Final Verdict

Now, is this point a maximum or a minimum? We turn to the Second Derivative Test. Differentiating $P^{\prime }(x)$ again, we obtain:

When we evaluate this at our critical point $x=0$, we get $P^{\prime \prime }(0)=2a_1$. Since $a_1>0$, the second derivative is positive.

A positive second derivative means the curve is concave upwards—it is shaped like a smile. A smile has a bottom, not a top. Thus, $x=0$ is a local minimum.

Because it is the only critical point, the function has only one minimum and no maximum. You have successfully navigated the complexity and found the truth hidden in the symmetry. Keep this analytical mindset, and no polynomial will ever intimidate you again.