Analyzing Function Properties and Derivatives

The problem provides information about a function \(f(x)\) and its derivatives, along with a derived function \(h(x)\). We are given:

• \(f'(x) = g(x)\)
• \(f''(x) = -f(x)\)
• \(h(x) = \{f(x)\}^2 + \{g(x)\}^2\)

We need to evaluate the correctness of two statements regarding \(h(x)\):

1. \(h'(3) = 0\)
2. \(h(1) = h(2)\)

Step-by-Step Evaluation of Statement 1: \(h'(3) = 0\)

To check if \(h'(3) = 0\), we first need to find the general expression for the derivative of \(h(x)\), which is \(h'(x)\).

The function \(h(x)\) is defined as the sum of squares of \(f(x)\) and \(g(x)\):

\(h(x) = \{f(x)\}^2 + \{g(x)\}^2\)

Using the chain rule for differentiation, the derivative \(h'(x)\) is:

\(h'(x) = \frac{d}{dx} [\{f(x)\}^2] + \frac{d}{dx} [\{g(x)\}^2]\)

\(h'(x) = 2f(x) \cdot f'(x) + 2g(x) \cdot g'(x)\)

Now, we can use the given information to substitute \(f'(x)\) and \(g'(x)\).

We are given \(f'(x) = g(x)\). This is a direct substitution.

To find \(g'(x)\), we can differentiate the equation \(f'(x) = g(x)\) with respect to \(x\):

\(\frac{d}{dx} [f'(x)] = \frac{d}{dx} [g(x)]\)

\(f''(x) = g'(x)\)

We are also given that \(f''(x) = -f(x)\). Therefore, we can conclude that \(g'(x) = -f(x)\).

Now, substitute these expressions for \(f'(x)\) and \(g'(x)\) into the equation for \(h'(x)\):

\(h'(x) = 2f(x) \cdot (g(x)) + 2g(x) \cdot (-f(x))\)

\(h'(x) = 2f(x)g(x) - 2g(x)f(x)\)

\(h'(x) = 0\)

The derivative \(h'(x)\) is 0 for all values of \(x\). This means that \(h(x)\) is a constant function. Since \(h'(x) = 0\) for all \(x\), it is specifically true for \(x = 3\).

Therefore, \(h'(3) = 0\).

Statement 1 is correct.

Step-by-Step Evaluation of Statement 2: \(h(1) = h(2)\)

From the analysis of Statement 1, we found that \(h'(x) = 0\) for all values of \(x\). A function whose derivative is zero everywhere on an interval is a constant function on that interval.

Since \(h'(x) = 0\) for all \(x\), \(h(x)\) is a constant function. This means that the value of \(h(x)\) is the same for any value of \(x\).

Therefore, \(h(1)\) must be equal to \(h(2)\).

Statement 2 is correct.

Conclusion on the Statements

Both statement 1 (\(h'(3) = 0\)) and statement 2 (\(h(1) = h(2)\)) are correct based on the given conditions \(f'(x) = g(x)\) and \(f''(x) = -f(x)\).

Summary of Findings

• By differentiating \(h(x) = \{f(x)\}^2 + \{g(x)\}^2\) and using the given derivative relationships, we found that \(h'(x) = 0\).
• Since \(h'(x) = 0\) for all \(x\), Statement 1, \(h'(3) = 0\), is true.
• A function with a derivative of zero is a constant function, so \(h(x)\) is constant.
• Since \(h(x)\) is constant, its value at \(x=1\) is the same as its value at \(x=2\). Thus, Statement 2, \(h(1) = h(2)\), is true.

Revision Table: Function Properties

Additional Information: Constant Functions and Derivatives

A function \(F(x)\) is considered a constant function over an interval if its value does not change for any \(x\) in that interval. A key property in calculus is that if the derivative of a function \(F'(x)\) is equal to zero for all \(x\) in an interval, then the function \(F(x)\) is a constant function on that interval. Conversely, the derivative of any constant function is always zero.

In this problem, we found that \(h'(x) = 0\) for all \(x\). This immediately tells us that \(h(x)\) is a constant function. If a function is constant, its value at any point \(a\) is the same as its value at any other point \(b\). This is why \(h(1) = h(2)\) is true.

The given conditions \(f''(x) = -f(x)\) are characteristic of functions like \(\sin(x)\) and \(\cos(x)\). For instance, if \(f(x) = \sin(x)\), then \(f'(x) = \cos(x)\) (so \(g(x) = \cos(x)\)) and \(f''(x) = -\sin(x)\). If \(f(x) = \cos(x)\), then \(f'(x) = -\sin(x)\) (so \(g(x) = -\sin(x)\)) and \(f''(x) = -\cos(x)\). Let's check if \(h(x)\) is constant for these: If \(f(x) = \sin(x)\), then \(g(x) = \cos(x)\). \(h(x) = \{\sin(x)\}^2 + \{\cos(x)\}^2 = \sin^2(x) + \cos^2(x) = 1\). This is a constant function. If \(f(x) = \cos(x)\), then \(g(x) = -\sin(x)\). \(h(x) = \{\cos(x)\}^2 + \{-\sin(x)\}^2 = \cos^2(x) + \sin^2(x) = 1\). This is also a constant function.

This confirms our finding that \(h(x)\) must be a constant function under the given conditions, regardless of the specific form of \(f(x)\) (as long as it satisfies the derivative relationships).