Animated Solution for Mathematics - Differentiation: Comprehension Passage

Welcome, fellow traveler of the mathematical landscape. Today, we are dissecting a function that might look like a standard rational expression at first glance, but beneath its surface lies a beautiful interplay of calculus, algebra, and logic.

We are working with the function:

where $0<a<2$. Before we even touch a derivative, let us appreciate the domain.

As we discussed, the denominator $x^2+ax+1$ has a discriminant $D=a^2-4$. Because $a$ is trapped between $0$ and $2$, $a^2$ is strictly less than $4$, meaning $D<0$.

This is our first victory: the denominator is never zero. The function is smooth, continuous, and well-defined across the entire real number line.

Unveiling the Monotonicity

Now, we seek the derivative $f^{\prime }(x)$. We apply the quotient rule: $\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$.

It is a tedious process, but if you keep your terms organized, you will see the magic happen. After differentiating the numerator and denominator and performing the subtraction, the expression simplifies beautifully to:

Look at this result. The denominator is a square, so it is always positive. The constant $2a$ is positive.

The entire sign of the derivative is dictated by $(x^2-1)$. This is the key!

When $x^2<1$ (i.e., $x\in (-1,1)$), the derivative is negative, meaning the function is decreasing. When $x^2>1$, the function is increasing. This simple observation unlocks the entire behavior of the curve.

The problem asks us to evaluate the second derivative at $x=1$ and $x=-1$. This is where we test our algebraic stamina.

By differentiating $f^{\prime }(x)$ again, we arrive at:

Evaluating this at $x=1$ gives us $\frac{4a}{(2+a{)}^2}$, and at $x=-1$, we get $-\frac{4a}{(2-a{)}^2}$.

When we check the relation $(2+a{)}^2{f}^{\prime \prime }(1)+(2-a{)}^2{f}^{\prime \prime }(-1)$, we see the terms $4a$ and $-4a$ cancel out perfectly to zero. It is a moment of pure mathematical elegance—the kind that reminds us why we solve these problems.

Furthermore, since $f^{\prime }(1)=0$ and $f^{\prime \prime }(1)>0$, the Second Derivative Test confirms that $x=1$ is a local minimum.

The Final Frontier

Finally, we encounter the function:

Many students panic when they see an integral with a variable upper limit. Do not fear! The Leibniz Integral Rule is your best friend here.

We differentiate $g(x)$ with respect to $x$:

This simplifies to:

Now, analyze the sign. $e^x$ is always positive, and $1+e^{2x}$ is always positive. The sign of $g^{\prime }(x)$ depends entirely on $f^{\prime }(e^x)$.

If $x<0$, then $e^x<1$, which means $f^{\prime }(e^x)<0$. If $x>0$, then $e^x>1$, which means $f^{\prime }(e^x)>0$.

Thus, $g^{\prime }(x)$ is negative on $(-\infty ,0)$ and positive on $(0,\infty )$. We have successfully mapped the entire behavior of the function. You have navigated the algebra, mastered the calculus, and arrived at the truth. Keep this confidence with you—you are ready for the next challenge.