Question

Let h(x)=e−x+kx, 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?

Answer 1

Answer:

a) for k≤0 , h has no critical point

b) for k>0 , h has a critical point

c) for k=0 , has a horizontal asymptote

Step-by-step explanation:

for the function

h(x)=e^(−x)+k

h has a critical point when the first derivative is =0 or is undefined. Since e^(−x) and k*x are continuos functions for all x then the second case is discarded. Then

dh/dx = -e^(−x)+k = 0

k = e^(−x)

x = ln (1/k)

since ln (1/k) should be possitive then k should be >0 . Thus h(x) has a critical point when k>0 and do not have any when  k≤0

h has a horizontal asymptote when

lim h(x)=a when x→∞ (or -∞)

then

when x→∞, lim h(x)= lim e^(−x)+k*x = lim e^(−x) + k* lim x = 0 + k*∞ = ∞

on the other hand , when k=0 , lim h(x)= lim e^(−x)= 0 , then h has a horizontal  asymptote for k=0

for x→(-∞) , e^(-x) rises exponentially , thus there is no k such that h has an horizontal asymptote when x→(-∞)

Answer 2

a). The function h(x) has no critical point for  k≤0

b). The function h(x) has a critical point for   k>0

c). The function h(x )has a horizontal asymptote for k = 0.

Critical points:

The given function is,

                   $h(x)=e^{-x} +kx$

Function h(x) has a critical point when the first derivative  =0

                $\frac{dh}{dx}=-e^{-x} +k=0\\\\k=e^{-x} \\\\x=ln(\frac{1}{k} )$

Thus h(x) has a critical point when k>0 and do not have any critical points when  k≤0

The function h(x)  has a horizontal asymptote when,

        $\lim_{x \to \infty} h(x)=0$

   $\lim_{x \to \infty} (e^{-x}+kx)=k*\infty=\infty$

When k = 0, $\lim_{x \to \infty} (e^{-x}+kx)=0*\infty=0$

Therefore, function h(x) has a horizontal  asymptote for k=0

Learn more about the asymptotes here:

https://brainly.com/question/1851758