Question

The range of \(\mathrm{f}(\mathrm{x})=6^{\mathrm{x}}+3^{\mathrm{x}}+6^{-\mathrm{x}}+3^{-\mathrm{x}}+2\) is subset of \(\ldots\) (a) \([-\infty,-2)\) (b) \((-2,3)\) (c) \((6, \infty)\) (d) \([6, \infty)\)

Short answer

The range of the function \(f(x)=6^{x}+3^{x}+6^{-x}+3^{-x}+2\) is a subset of \((6, \infty)\), because the values of the function are increasing for all real values of \(x\neq 0\) and its minimum value is 2 when \(x=0\).

Step-by-step solution

Identifying minimum point

The function is a sum of exponential terms and is positive for any real value of x. Hence, the sum is also positive for any real value of \( x \). To find the minimum point of this function, we should find the value of \( x \) that makes the function as small as possible.

Differentiating The Function

Differentiating the function will provide us with its gradient and might help to find points of local extrema if they exist. The first derivative of \(\mathrm{f(x)}=6^{\mathrm{x}}+3^{\mathrm{x}}+6^{-\mathrm{x}}+3^{-\mathrm{x}}+2\) is: \[ f'(x)= 6^{x} \ln 6 + 3^{x} \ln 3 - 6^{-x} \ln 6 - 3^{-x} \ln 3\]

Setting The First Derivative To Zero

Points of local extrema (minimum or maximum) can usually be found by setting the first derivative to zero. However, solving \(0= 6^{x} \ln 6 + 3^{x} \ln 3 - 6^{-x} \ln 6 - 3^{-x} \ln 3\) for all real x is very complex and doesn't produce clear solutions that would help to establish the range.

Analyzing The Function

Upon examination, it can be seen that the function evaluates to 2 when x equals 0. The exponentials \(6^x\) and \(3^x\) grow much faster for positive x than the inverses \(6^{-x}\) and \(3^{-x}\), and conversely when x is negative. This indicates that the function values increase in both the positive and negative directions of x, from a minimum of 2 at x=0.

Choosing the Correct Option

As established in Step 4, the function is increasing for all values of \(x\neq 0\), and its minimum value is 2 when \(x=0\). Therefore, the function values will always be greater than or equal to 2. The only subset that accounts for all values greater than 2, is the interval \((6, \infty)\), hence solution (c).

Key concepts

Exponential Function Analysis

Understanding exponential functions, like the one presented in the exercise, is vital for grasping a variety of mathematical and real-world concepts. An exponential function is of the form \( f(x) = a^x \) where \( a \) is a positive real number, and \( a eq 1 \). It is characterized by a rapid increase or decrease as \( x \) moves away from zero.

Analysis of these functions involves looking at their growth rates, asymptotes, and end behavior. In our exercise, the function \( f(x) = 6^{x} + 3^{x} + 6^{-x} + 3^{-x} + 2 \) contains both growing and decaying exponential terms as \( x \) increases in either direction—positive or negative. The plus two (\( +2 \) ) at the end functions as a vertical shift, moving the entire graph of the function up by two units.

Growth Rate Comparison

For the growing parts \( 6^{x} \) and \( 3^{x} \), as \( x \) becomes larger, \( 6^{x} \) grows faster than \( 3^{x} \) due to the larger base. Conversely, for the decaying parts \( 6^{-x} \) and \( 3^{-x} \), \( 6^{-x} \) approaches zero faster than \( 3^{-x} \) as \( x \) becomes more negative. While analyzing exponential functions, we always consider these rates to predict the function's behavior far from the origin.

Finding Range of Functions

The range of a function refers to the set of all possible output values (y-values) it can produce. Finding the range is important for understanding the limits within which the function operates. There are different methods to find the range of functions, including graphical observation, transformation techniques, and calculus-based methods.

The range can be particularly challenging to find for more complex functions that involve sums of different terms, as in our exercise. One approach is to graph the function and observe its behavior, but for more precise analysis, calculus often provides the tools needed.

Utilizing Derivatives

By using derivatives, we can determine the increasing or decreasing nature of the function, and identify local and global extrema, which can guide us to the range. In our textbook exercise, it is noticed that for \( x = 0 \), the function simplifies to a minimum value of 2. From there, the function values can increase without bound as \( x \) increases or decreases, indicating that the range extends from \( y = 2 \) to infinity. This understanding aligns with the solution that the range is a subset of \( (6, \infty) \) and excludes values less than 2.

Setting Derivatives to Zero

A fundamental technique in calculus is setting the derivative of a function to zero to find its critical points. These points are where the function's slope is zero, indicating potential local maxima, minima, or points of inflection.

When we set the derivative to zero, we're searching for values of \( x \) where the function's rate of change is zero, which corresponds to a horizontal tangent on the graph. These \( x \) values can then be substituted back into the original function to determine the corresponding \( y \) values, establishing potential extrema.

Exploring Complex Derivatives

In the case of our example function, setting the derivative \( f'(x)= 6^{x} \ln 6 + 3^{x} \ln 3 - 6^{-x} \ln 6 - 3^{-x} \ln 3 \) to zero results in a complex equation. However, through critical analysis—as employed in the textbook solution—it's clear that solving this equation is not straightforward, and other methods like examining the function’s behavior at \( x = 0 \) or considering the nature of exponential growth and decay are more effective in determining the range.