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Identify the inner and outer functions in the composition \(\left(x^{2}+10\right)^{-5}.\)

Short Answer

Expert verified
Answer: The inner function is \(x^2 + 10\), and the outer function is raising to the power of \(-5\).

Step by step solution

01

Identify the inner function

The inner function is the one inside the parentheses. In this case, the inner function is \(x^2 + 10\).
02

Identify the outer function

The outer function is the one applied to the result of the inner function. In this case, the outer function is raising to the power of \(-5\), which can be written as \(f(u) = u^{-5}\).
03

Write the composition as a combination of inner and outer functions

In order to show the composition explicitly as a combination of inner and outer functions, we can write it as \(f(g(x))\), where \(g(x) = x^2 + 10\) is the inner function and \(f(u) = u^{-5}\) is the outer function. So, the composition is \(f(g(x)) = \left(g(x)\right)^{-5} = \left(x^2 + 10\right)^{-5}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inner Function
When we talk about function composition, the inner function is like the first flavor of a two-layered cake you taste. In the composition \( (x^2 + 10)^{-5} \), the inner function is gracefully nestled within the parentheses, ready to be evaluated first. It is represented by \( g(x) = x^2 + 10 \). Think of this inner function as the starting point. It gets things moving, transforming the input, which in this case is 'x', with a neat square and a friendly addition of 10. Understanding the inner function is crucial because it sets up the value that the outer function will then take over and transform.
Outer Function
If the inner function is the first layer of our cake, the outer function is the top layer that completes the composition. Focusing back on our example, the outer function hones in on whatever the inner function outputs and does its own thing to it, which in this scenario means it takes that result and raises it to the power of \(-5\), represented as \( f(u) = u^{-5} \). This flip in sign serves as a reminder that this function will essentially flip our previous result by taking a negative power, spreading a new perspective over our initial savory square and addition. In simpler terms, you apply this outer function to the result from the inner function, and voila, you have your composite function.
Composition of Functions
Merging our inner and outer functions, we step into the broader, swirling world of composition of functions. You've probably been to a concert where a band plays a mashup of two songs - this is the mathematical equivalent. By taking the output from our inner function, \( g(x) \), and putting it through the churning mechanism of our outer function, \( f(u) \), we create this magnificent piece, the composite function, \( f(g(x)) = (g(x))^{-5} = (x^2 + 10)^{-5} \). It's a dance of cooperation where one functions output waltzes right into the embrace of the other function to create something new and complex.
Exponential Functions
Lastly, let's not overlook the star attraction in our example, the power of exponential functions. These functions are like the roller coasters of the math world, taking simple bases and raising them to dizzy heights (or plunging depths) with their exponents. In our case, the \( u^{-5} \) indicates an inverse exponential relationship. Typically, when we see an expression like \( y = b^x \), 'b' is the base and 'x' is the exponent. Exponential functions grow (or decay) extraordinarily fast - they're not linear or even polynomial. As they chart their courses on graphs, they curve upwards or downwards with gusto, often used in calculations with exponential growth or decay, such as interest rates or populations. Exponential functions are all about the power - they take whatever value you give them and elevate it, quite literally, to the next level.

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Most popular questions from this chapter

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. Find \(d y / d t\), the velocity of the mass. Assume \(k\) and \(m\) are constant. b. How would the velocity be affected if the experiment were repeated with four times the mass on the end of the spring? c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness ( \(k\) is increased by a factor of 4 )? d. Assume y has units of meters, \(t\) has units of seconds, \(m\) has units of kg, and \(k\) has units of \(\mathrm{kg} / \mathrm{s}^{2} .\) Show that the units of the velocity in part (a) are consistent.

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u \end{array}\right.$$ a. Show that \(\lim _{y \rightarrow u} H(v)=0\) b. For any value of \(u,\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u)$$ c. Show that $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right)$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\)

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. The period \(T\) is the time required by the mass to complete one oscillation. Show that \(T=2 \pi \sqrt{\frac{m}{k}}\) b. Assume \(k\) is constant and calculate \(\frac{d T}{d m}\) c. Give a physical explanation of why \(\frac{d T}{d m}\) is positive.

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