MonotonicityHard
Question
The values of p for which the function f(x) = 
x5 - 3x + ln 5 decreases for all real x is
x5 - 3x + ln 5 decreases for all real x isOptions
A.(- ∞, ∞)
B.
υ (1, ∞)
υ (1, ∞)C.
υ (2, ∞)
υ (2, ∞)D.[1, ∞)
Solution
f(x) = 
x5 - 3x + ln 5
f′(x) =
5x4 - 3 < 0 ∀ x ∈ R
⇒
- 1 ≤ 0
If - 4 ≤ p < 1 then
⇒
≤ 1 - p ⇒ p + 4 ≤ 1 - 2p + p2
⇒ p2 - 3p - 3 ≥ 0
⇒ p ≤
≤ p
⇒ p∈
If p > 1 then
≥ 1 - p
⇒ Always true for p > 1
⇒ p∈
υ (1, ∞)
x5 - 3x + ln 5f′(x) =
5x4 - 3 < 0 ∀ x ∈ R⇒
- 1 ≤ 0If - 4 ≤ p < 1 then
⇒
≤ 1 - p ⇒ p + 4 ≤ 1 - 2p + p2⇒ p2 - 3p - 3 ≥ 0
⇒ p ≤
≤ p⇒ p∈
If p > 1 then
≥ 1 - p⇒ Always true for p > 1
⇒ p∈
υ (1, ∞)Create a free account to view solution
View Solution FreeMore Monotonicity Questions
For the function f(x) = x3 − 6x2 − 36x + 7 which of the following statement is false...In the following increasing function is -...The function f(x) = cos x - 2px is monotonically decreasing for...The interval in which the function f(x) = xe4−x decreases is...Consider the function f(x) = then the number of points in (0, 1) where the derivative f′(x) vanishes, is -...