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Chapter 4: Problem 30

Let \(h(x)=e^{-x}+k x,\) where \(k\) is any constant. For what value(s) of \(k\) does \(h\) have (a) No critical points? (b) One critical point? (c) A horizontal asymptote?

Short Answer

Expert verified
(a) \( k \leq 0 \), (b) \( k > 0 \), (c) \( k = 0 \).

Step by step solution

01

Differentiate h(x)

To find the critical points, we first need the derivative of the function. The function is given by \( h(x) = e^{-x} + kx \). The derivative \( h'(x) \) is found by differentiating each term separately. The derivative of \( e^{-x} \) is \( -e^{-x} \), and the derivative of \( kx \) is \( k \). Thus, \( h'(x) = -e^{-x} + k \).
02

Solve for Critical Points

Critical points occur where \( h'(x) = 0 \). Setting the derivative equal to zero gives \( -e^{-x} + k = 0 \). Rearranging gives \( e^{-x} = k \). To find \( x \), take the natural logarithm of both sides to obtain \( x = -\ln(k) \). Critical points exist if \( k > 0 \) (as the exponential function is positive).
03

Analyse Values for k

We have the equation \( x = -\ln(k) \) for critical points, which is defined only for \( k > 0 \). Therefore:(a) For no critical points, \( k \leq 0 \) as the equation \( x = -\ln(k) \) is not defined.(b) For one critical point, \( k > 0 \) as \( x \) is defined only when \( k \) is positive.
04

Check for Horizontal Asymptotes

A horizontal asymptote occurs when the function approaches a constant value as \( x \to \infty \) or \( x \to -\infty \). As \( x \to \infty \), \( e^{-x} \rightarrow 0 \), so the term \( kx \) determines this behavior. There is no horizontal asymptote since \( kx \) tends toward infinity for \( k eq 0 \). For \( k = 0 \), \( h(x) = e^{-x} \) which approaches \( 0 \), thus having a horizontal asymptote at \( y = 0 \).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing at any given point. In simpler terms, if you want to know how steep a curve is at a particular point, you'd need to find the derivative of the function representing that curve. Differentiation allows us to obtain this derivative.

For the function given in the exercise, \( h(x) = e^{-x} + kx \), differentiation is used to determine its derivative, \( h'(x) \).
  • The process involves differentiating each term separately.
  • For the term \( e^{-x} \), the derivative is \( -e^{-x} \) because the chain rule applies here.
  • The derivative of \( kx \) is simply \( k \), since the derivative of \( x \) with respect to \( x \) is 1, and \( k \) is a constant multiplier.
So, the derivative \( h'(x) = -e^{-x} + k \). This derivative is crucial to finding key characteristics of the function, like where it has critical points or whether it possesses a horizontal asymptote.
Critical Points
Critical points are where a function's derivative is zero or undefined. These points are significant because they can indicate where the function has a relative maximum, relative minimum, or a point of inflection. To find these, set the derivative of the function equal to zero and solve for the variable.

In the exercise, we start with the derivative \( h'(x) = -e^{-x} + k \). Setting this equal to zero:
  • \( -e^{-x} + k = 0 \)
  • Rearrange to find \( e^{-x} = k \)
  • Taking the natural logarithm gives \( x = -\ln(k) \)
Critical points will exist only if \( k > 0 \), as we are taking the logarithm of \( k \), which is only defined for positive values of \( k \). Hence:
  • No critical points: If \( k \leq 0 \), the equation is not valid.
  • One critical point: If \( k > 0 \), the point \( x = -\ln(k) \) is defined.
Managing critical points helps in understanding the behavior of graphs and their turning points.
Horizontal Asymptotes
Horizontal Asymptotes describe the behavior of a function as the input approaches positive or negative infinity. Essentially, if as \( x \to \pm \infty \), the function approaches some constant value \( L \), then \( y = L \) is a horizontal asymptote.

For the function \( h(x) = e^{-x} + kx \), we analyze \( x \to \infty \):
  • For \( x \to \infty \), \( e^{-x} \) approaches 0, as exponential decay means it tends to zero.
  • The term \( kx \) will grow indefinitely large when \( k \) is non-zero, as multiplying a linear function by any non-zero constant results in a term tending to \( \pm \infty \).
Consequently, \( h(x) = e^{-x} + kx \) has no horizontal asymptote when \(k eq 0\).
  • If \( k = 0 \), \( h(x) = e^{-x} \) and approaches 0 as \( x \to \infty \),
Thus, the function has a horizontal asymptote at \( y = 0 \) when \( k = 0 \). Understanding asymptotes provides insights into the long-term trends of functions.

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Most popular questions from this chapter

The Lennard-Jones model predicts the potential energy \(V(r)\) of a two-atom molecule as a function of the distance \(r\) between the atoms to be $$V(r)=\frac{A}{r^{12}}-\frac{B}{r^{6}}, \quad r>0$$ where \(A\) and \(B\) are positive constants. (a) Evaluate \(\lim _{r \rightarrow 0^{+}} V(r),\) and interpret your answer. (b) Find the critical point of \(V(r) .\) Is it a local maximum or local minimum? (c) The inter-atomic force is given by \(F(r)=-V^{\prime}(r)\) At what distance \(r\) is the inter-atomic force zero? (This is called the equilibrium size of the molecule.) (d) Describe how the parameters \(A\) and \(B\) affect the equilibrium size of the molecule.

Are the statements in Problems true or false for a function \(f\) whose domain is all real numbers? If a statement is true,explain how you know. If a statement is false, give a counterexample. If \(x=p\) is not a critical point of \(f,\) then \(x=p\) is not a local maximum of \(f\).

Are the statements in Problems true or false for a function \(f\) whose domain is all real numbers? If a statement is true,explain how you know. If a statement is false, give a counterexample. A local maximum of \(f\) occurs at a point where $$ f^{\prime}(x)=0. $$

Find formulas for the functions described. A function of the form \(y=b x e^{-a x}\) with a local maximum at (3,6).

If \(a>0, b>0,\) show that \(f(x)=a\left(1-e^{-b x}\right)\) is everywhere increasing and everywhere concave down.
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