MonotonicityHard
Question
If a < 0, the function f(x) = eax + e-ax is monotonically decreasing for all values of x, where-
Options
A.x > 0
B.x < 0
C.x > 1
D.x < 1
Solution
f(x) = eax + e-ax
⇒ f′(x) = a [eax - e-ax]
= 2a
= 2a2x
Now f′(x) < 0 ⇒ x < 0
Hence, f(x) is decreasing for x < 0.
f(x) = eax + e-ax
⇒ f′(x) = a [eax - e-ax]
= 2a
= 2a2x
Now f′(x) < 0 ⇒ x < 0
Hence, f(x) is decreasing for x < 0.
⇒ f′(x) = a [eax - e-ax]
= 2a
= 2a2x
Now f′(x) < 0 ⇒ x < 0
Hence, f(x) is decreasing for x < 0.
f(x) = eax + e-ax
⇒ f′(x) = a [eax - e-ax]
= 2a
= 2a2x
Now f′(x) < 0 ⇒ x < 0
Hence, f(x) is decreasing for x < 0.
Create a free account to view solution
View Solution Free