Verified Answer :
- Derive \(f(x) = 5e^{\cos 2x}\) using the chain rule to get \(f'(x) = -10\sin(2x)e^{\cos 2x}\).
- Derive the linear function \(g(x) = \log_3(5^{4x})\) to get the constant slope \(g'(x) = \frac{4\ln 5}{\ln 3}\).
- Solve the equation \(f'(x) = g'(x)\) by finding the root of \(h(x) = f'(x) - g'(x) = 0\) within \([-1, 1]\).
- The solution is \(x \approx \boxed{-0.1114}\).
Explanation
- Derivatives of the Functions
To find the values of \(x \in [-1, 1]\) where the functions \(f(x) = 5e^{\cos 2x}\) and \(g(x) = \log_3(5^{4x})\) have parallel tangent lines, we must find the values of \(x\) such that their derivatives are equal, that is, \(f'(x) = g'(x)\).
First, consider the function \(f(x) = 5e^{\cos 2x}\). Applying the chain rule, we differentiate with respect to \(x\):
\[f'(x) = 5 \cdot e^{\cos 2x} \cdot \frac{d}{dx}(\cos 2x) = 5 \cdot e^{\cos 2x} \cdot (-\sin 2x \cdot 2) = -10 \sin(2x) e^{\cos 2x}\]
Next, consider the function \(g(x) = \log_3(5^{4x})\). We can simplify this using the property of logarithms \(\log_b(a^c) = c \log_b(a)\) and the change-of-base formula \(\log_b(a) = \frac{\ln a}{\ln b}\):
\[g(x) = 4x \log_3(5) = 4x \frac{\ln 5}{\ln 3}\]
Since this is a linear function of the form \(g(x) = mx + c\), its derivative is simply the coefficient of \(x\):
\[g'(x) = \frac{4 \ln 5}{\ln 3}\]
Using numerical approximation, \(g'(x) \approx \frac{4 \cdot 1.6094}{1.0986} \approx 5.8599\).
- Finding the Intersection Point
We set the derivatives equal to each other to find the values of \(x\) where the tangent lines are parallel:
\[-10 \sin(2x) e^{\cos 2x} = \frac{4 \ln 5}{\ln 3}\]
We define the function \(h(x) = -10 \sin(2x) e^{\cos 2x} - \frac{4 \ln 5}{\ln 3}\) and seek its roots in the interval \([-1, 1]\). We have already computed the value of the constant \(\frac{4 \ln 5}{\ln 3} \approx 5.8599\).
The equation is:
\[-10 \sin(2x) e^{\cos 2x} = 5.8599\]
We solve this for \(x\) in the interval \([-1, 1]\) using numerical methods. The calculations show there is a unique root at \(x \approx -0.1114\). To verify this:
If \(x = -0.1114\), \(2x \approx -0.2228\).
Then, \(-10 \sin(-0.2228) e^{\cos(-0.2228)} \approx -10 \cdot (-0.2209) \cdot e^{0.9753} \approx 2.209 \cdot 2.6519 \approx 5.858\).
This confirms the root is indeed near \(-0.1114\).
- Final Answer
The tangent lines to \(f(x) = 5e^{\cos 2x}\) and \(g(x) = \log_3(5^{4x})\) are parallel when their slopes are equal, which occurs at \(x \approx -0.1114\). This point lies within the specified interval of \([-1, 1]\). The calculation confirms this is the only solution in that range. The final answer is approximately \(\boxed{-0.1114}\).
Examples
In structural engineering, the concept of parallel tangent lines is essential when designing beams or surfaces that need to maintain a constant relative angle or orientation at specific points. For instance, if you are designing a suspension bridge cable curve and a secondary support structure curve, ensuring their tangent lines are parallel at certain intervals can help maintain uniform stress distribution across the connections. Determining these points using calculus, as we did with the functions \(f(x)\) and \(g(x)\), allows engineers to predict these critical alignment points with precision.