Q. \(f(1+x)+f(1-x)=0\), \(f(-x)=f(x)\), \(2^{x}-1\).

Q: What does f(1+x)+f(1-x)=0 express?

It says f is odd with respect to the point 1: define g(x)=f(1+x); then g(-x)=-g(x). So values symmetric about 1 are negatives.

Q: Can a nonzero function satisfy both conditions?

Yes. For example f(x) = \cos\left(\frac{\pi x}{2}\right) is even and gives f(1+x) + f(1-x) = 0. So nontrivial solutions exist.

Q: Why is the period 4 conclusion valid?

From f(1+x)+f(1-x)=0 get f(2+x)=-f(x). Applying again gives f(4+x)=f(x), so f is 4-periodic.

Q: Does f(x)=2^x-1 satisfy the conditions?

Does f(x)=2^x-1 satisfy the conditions?

Q: What are immediate values at integers?

From x=0 in f(1+x)+f(1-x)=0 get 2f(1)=0\Rightarrow f(1)=0. Evenness gives f(-1)=0. Periodicity implies many other integer relations.

Q: How to check a candidate function quickly?

Verify evenness: test f(-x)=f(x). Then check the shift-odd condition: compute f(1+x)+f(1-x) and see if it vanishes for all x (or check Fourier coefficients fit the odd-index cosine pattern).

Answer

The given functional equations describe properties of a function \(f(x)\):

1. \(f(-x)=f(x)\) indicates that the function is even (symmetric about the \(y\)-axis).

2. \(f(1+x)+f(1-x)=0\) implies the function is odd with respect to the point \(x=1\), which also means \(f(x)\) is periodic with a period of \(4\).

The expression \(2^x-1\) is not a solution to these equations. Evaluating \(2^x-1\):

– Evenness check: \(f(-1)=2^{-1}-1=-\tfrac{1}{2}\), while \(f(1)=2^1-1=1\). Since \(f(-1)\neq f(1)\), it is not even.

– Symmetry check: \(f(1+0)+f(1-0)=1+1=2\neq 0\).

Conclusion: The function \(2^x-1\) does not satisfy the specified conditions. A valid example that satisfies both conditions is \(f(x)=\cos\!\left(\frac{\pi x}{2}\right)\).

Detailed Explanation

Analysis of the Functional Equations for (f(x))

The problem provides two functional equations and asks us to analyze them or evaluate a specific function \(2^x – 1\) against them. The equations are:

\(f(1 + x) + f(1 – x) = 0\)
\(f(-x) = f(x)\)

Step 1: Understand the Symmetry Conditions

Equation 1: \(f(1 + x) + f(1 – x) = 0\) can be rewritten as \(f(1 + x) = -f(1 – x)\). This indicates that the function is odd with respect to the point \(x = 1\). This means the graph of the function has rotational symmetry of 180 degrees around the point \((1,0)\).

Equation 2: \(f(-x) = f(x)\) defines an even function. This means the graph is symmetric with respect to the y-axis.

Step 2: Derive Periodicity

To see how these two symmetries interact, we can perform a substitution. From Equation 1, let \(x\) be replaced by \(x + 1\):

\(f(1 + (x + 1)) + f(1 – (x + 1)) = 0\)

That is, \(f(x + 2) + f(-x) = 0\).

Using Equation 2 (\(f(-x) = f(x)\)), we substitute \(f(x)\) for \(f(-x)\):

\(f(x + 2) + f(x) = 0\)

\(f(x + 2) = -f(x)\).

Now, to find the period, apply this result twice:

\(f(x + 4) = f((x + 2) + 2) = -f(x + 2) = -(-f(x)) = f(x)\).

Conclusion: Any function satisfying both conditions must be periodic with period 4.

Step 3: Evaluate \(f(x) = 2^x – 1\)

Now we check if the specific expression \(2^x – 1\) satisfies these properties.

Check Evenness:

\(f(-x) = 2^{-x} – 1 = \dfrac{1}{2^x} – 1\).

Since \(\dfrac{1}{2^x} – 1 \neq 2^x – 1\) for general \(x\), the function is not even. Thus, it fails Equation 2.

Check Symmetry about \(x=1\):

\(f(1+x) + f(1-x) = (2^{1+x} – 1) + (2^{1-x} – 1) = 2\cdot 2^x + \dfrac{2}{2^x} – 2\).

This expression does not equal zero for all \(x\). For example, if \(x=0\), we get \(2\cdot 1 + 2\cdot 1 – 2 = 2 \neq 0\). Thus, it fails Equation 1.

Final Conclusion

The function \(f(x) = 2^x – 1\) does not satisfy the given functional equations. Functions that do satisfy these conditions are typically trigonometric, such as \(f(x) = \cos\!\left(\dfrac{\pi x}{2}\right)\).

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FAQs

What does the condition \(f(-x)=f(x)\) mean?

It means f is an even function: symmetric about 0. Graphically, values at ±x are equal. Analytically, an expansion uses only even powers or cosine terms in Fourier series.

What does \(f(1+x)+f(1-x)=0\) express?

It says f is odd with respect to the point 1: define \(g(x)=f(1+x)\); then \(g(-x)=-g(x)\). So values symmetric about 1 are negatives.

Can a nonzero function satisfy both conditions?

Yes. For example \(f(x) = \cos\left(\frac{\pi x}{2}\right)\) is even and gives \(f(1+x) + f(1-x) = 0\). So nontrivial solutions exist.

How do you find the general continuous solution?

Require evenness and the odd-about-1 condition; periodicity 4 follows. The general \(L^2\)/continuous solution is \(f(x) = \sum_{k \ge 0} a_k\cos\left(\frac{(2k+1)\pi x}{2}\right)\), with real coefficients \(a_k\).

Why is the period 4 conclusion valid?

From \(f(1+x)+f(1-x)=0\) get \(f(2+x)=-f(x)\). Applying again gives \(f(4+x)=f(x)\), so f is 4-periodic.

Does \(f(x)=2^x-1\) satisfy the conditions?

What are immediate values at integers?

From \(x=0\) in \(f(1+x)+f(1-x)=0\) get \(2f(1)=0\Rightarrow f(1)=0\). Evenness gives \(f(-1)=0\). Periodicity implies many other integer relations.

How to check a candidate function quickly?

Verify evenness: test \(f(-x)=f(x)\). Then check the shift-odd condition: compute \(f(1+x)+f(1-x)\) and see if it vanishes for all x (or check Fourier coefficients fit the odd-index cosine pattern).

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