Let f(x) be a function satisfying the condition f(-x) = f(x) for all real x. If f'(0) exists, find i

Q: Wait, why do we reject the negative root here? Or is it just that $f'(0)$ must be zero?

Great question! We aren't rejecting a root; we are deriving a necessary condition. By showing that $f'(0) = -f'(0)$, the only number that satisfies this equation is zero. It's a direct consequence of the symmetry constraint imposed on the derivative.

Q: Why can't we just use the power rule or chain rule here? It feels like overkill to use limits.

The power rule only works for specific functions like polynomials. This problem is about *any* differentiable even function. The limit definition is the universal tool that works for all functions, allowing us to prove this property for the entire class of even functions.

Q: I'm confused about the negative sign in the denominator. Why does it flip the sign of the whole limit?

Think of it as factoring out a constant. When you have $\frac{1}{-h}$, that is the same as $-\frac{1}{h}$. Since the limit is a linear operator, you can pull that constant $-1$ outside the limit entirely. That is exactly what transforms the expression into $-f'(0)$.

Question

Let $f(x)$ be a function satisfying the condition $f(-x) = f(x)$ for all real $x$. If $f'(0)$ exists, find its value.

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Why Even Functions Are Special

Imagine you are standing on a perfectly symmetrical bridge. If you look to your right, you see a path that curves upward; if you look to your left, you see the exact same curve mirrored.

This is the essence of an **even function**. Mathematically, we define this as $f(-x) = f(x)$ for all real $x$.

It means the function doesn't care if you step to the left or the right; the height of the graph remains identical. But what does this symmetry imply about the slope of the function at the very center, at $x=0$? Let's dive into the calculus to find out.

### The Microscope: Defining the Derivative

To understand the slope at $x=0$, we must use the formal definition of the derivative. We are looking for $f'(0)$, which tells us how the function behaves at the origin.

Using the first principle of derivatives, we write:

$$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h}$$

Geometrically, this limit represents the slope of the secant line connecting the point at $x=0$ to a nearby point at $x=h$. As $h$ shrinks toward zero, this secant line morphs into the tangent line. We are told the derivative exists, which means this limit is a well-defined, finite number.

### The Algebraic Dance: Symmetry in Action

Since the derivative exists, we can approach $x=0$ from the other side. Let's take a step of size $-h$ instead of $h$.

The derivative definition remains the same, but we substitute $-h$ for $h$:

$$f'(0) = \lim_{h 	o 0} \frac{f(-h) - f(0)}{-h}$$

Now, here is where the magic of the even function property kicks in. We know that $f(-h) = f(h)$.

Let's substitute this into our left-hand limit expression:

$$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{-h}$$

Notice the denominator? It contains a $-h$. We can pull that negative sign out of the limit entirely, as it is just a constant factor of $-1$:

$$f'(0) = - \left( \lim_{h 	o 0} \frac{f(h) - f(0)}{h} ight)$$

### The Elegant Conclusion

Look closely at the expression inside the parentheses. It is exactly the original definition of $f'(0)$ that we started with!

This leads us to the beautiful, simple equation:

$$f'(0) = -f'(0)$$

If we add $f'(0)$ to both sides, we get $2f'(0) = 0$, which forces **$f'(0) = 0$**.

This is not just an algebraic result; it is a profound geometric truth. For any smooth, even function, the tangent line at the axis of symmetry ($x=0$) must be perfectly horizontal.

If it were tilted even slightly, the symmetry would be broken—the function would be increasing on one side and decreasing on the other, which contradicts the definition of an even function. So, the next time you see an even function, remember: at the center, the slope is always zero.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x)$ be a function satisfying the condition $f (- x) = f (x)$ for all real $x$ . If $f^{'} (0)$ exists, find its value.

Enter Numerical Value:

Visualized Solution

Visualizing Symmetry

Defining the Derivative at $x = 0$

Exploring the Left-Hand Limit

Applying the Even Property

Relating the Two Expressions

Solving for $f^{'} (0)$

Conclusion and Intuition

The Sigma Insight: Differentiability of a Function

The Geometry of Symmetry

The Microscope

The Algebraic Dance

The Elegant Conclusion

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