The interval of increase of the function f(x)=x-ex+tan2π7 is

fx=x-ex+tan2π/7 fIx=1-exfx is increasingfIx>0 1>ex⇒ex<e0 ⇒x<0 ⇒x∈-∞,0

The interval of increase of the function f(x)=x-ex+tan2π7 is

fx=x-ex+tan2π/7 fIx=1-exfx is increasingfIx>0 1>ex⇒ex<e0 ⇒x<0 ⇒x∈-∞,0

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The interval of increase of the function $f (x) = x - e^{x} + \tan (\frac{2 π}{7}) i s$

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$(0, \infty)$

$(- \infty, 0)$

$(1, \infty)$

(–1, 1)

answer is A.

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Detailed Solution

$f (x) = x - e^{x} + \tan 2 π / 7 f^{I} (x) = 1 - e^{x}$

$f (x)$ is increasing

$f^{I} (x) > 0 1 > e^{x} \Rightarrow e^{x} < e^{0} \Rightarrow x < 0 \Rightarrow x \in (- \infty, 0)$

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The interval of increase of the function f(x)=x-ex+tan2π7 is

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The interval of increase of the function $f (x) = x - e^{x} + \tan (\frac{2 π}{7}) i s$