Answer:
a) The vertical asymptote(s) are x = − π / 2 and x = π / 2
b) The interval where the function is decreasing is (- π / 2, 0)
The interval where the function is increasing is (0, π / 2)
c) Minimum: P (0, 0)
d) The function is concave up in the the interval (- π / 2, π / 2)
Step-by-step explanation:
a) The vertical asymptote(s) are
lim (x→− π/2) (8x*tan(x)) = 8*(− π/2)*tan(− π/2) = -4 π*(- ∞) → - ∞
x = − π / 2
lim (x→ π/2) (8x*tan(x)) = 8*(π/2)*tan( π/2) = 4 π*(∞) → + ∞
x = π / 2
The horizontal asymptote(s) does not exist (DNE).
b) f'(x) = (8x*tan(x))' = Sin(x) + x*Cos(x)
If x = - π / 4 where − π / 2 < - π / 4 < 0 we have
f'(- π / 4) = Sin(- π / 4) + (- π / 4)*Cos(- π / 4)
f'(- π / 4) = -(√2/2) - (π / 4)(√2/2) = -(√2/2)((4 + π) / 4) < 0
The interval where the function is decreasing is (- π / 2, 0)
If x = π / 4 where 0 < π / 4 < π / 2 we have
f'( π / 4) = Sin( π / 4) + ( π / 4)*Cos( π / 4)
f'(- π / 4) = (√2/2) + (π / 4)(√2/2) = (√2/2)((4 + π) / 4) > 0
The interval where the function is increasing is (0, π / 2)
c) f(x) = 8x*tan(x)
f'(x) = (8x*tan(x))' = 0
⇒ tan(x) + x*Sec²(x) = 0
⇒ (Sin(x)/Cos(x)) + (x/Cos²(x)) = 0
⇒ Sin(x) + x*Cos(x) = 0 ⇒ x = 0
f(0) = 8*0*tan(0) = 0
Minimum: P (0, 0)
d) f''(x) = (Sin(x) + x*Cos(x))'
⇒ f''(x) = Cos(x) + Cos(x) - x*Sin(x)
⇒ f''(x) = 2*Cos(x) - x*Sin(x)
If x = 0 where − π / 2 < 0 < π / 2 we have
f''(0) = 2*Cos(0) - 0*Sin(0) = 2*(1) - 0 = 2 > 0
Then, the function is concave up in the the interval (- π / 2, π / 2)
