Question

Show that the function f(x) = tan x – 4x is strictly decreasing on [-π/3, π/3]​

Answer 1

f (x) = tanx – 4x ⇒ f ′(x) = sec2x – 4

$When \frac{ - \pi}{3} < x < \frac{\pi}{3} \\ \\ 1 < \sec(x) < 2 \\ \\ \therefore \: 1 < { \sec}^{2} x < 4 \rightarrow - 3 < ( { \sec}^{2} - 4) < 0 \\ \\ thus \: for \: \frac{ - \pi}{3} < x < \frac{\pi}{3} \: f(x) < 0 \\ \\ hence , \: f \: is \: strictly \: decreaing \: on \huge ( \small \frac{ - \pi}{3} ,\frac{\pi}{3} \huge)$

Answer 2

Answer:

Hello

Step-by-step explanation:

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