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→(-∞)
