f (x) = 8 x tan x, - π / 2 < x < π / 2 (a) Find the vertical asymptote (s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) x = Find the horizontal asymptote (s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) y = (b) Find the interval where the function is increasing. (Enter your answer using interval notation.) Find the interval where the function is decreasing. (Enter your answer using interval notation.) (c) Find the local maximum and minimum values. (If an answer does not exist, enter DNE.) local maximum value local minimum value (d) Find the interval where the function is concave up. (Enter your answer using interval notation.)

Question

Piper Delacruz

Mathematics

29 June, 10:15

f (x) = 8 x tan x, - π / 2 < x < π / 2 (a) Find the vertical asymptote (s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) x = Find the horizontal asymptote (s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) y = (b) Find the interval where the function is increasing. (Enter your answer using interval notation.) Find the interval where the function is decreasing. (Enter your answer using interval notation.) (c) Find the local maximum and minimum values. (If an answer does not exist, enter DNE.) local maximum value local minimum value (d) Find the interval where the function is concave up. (Enter your answer using interval notation.)

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Answers (1)

Know the Answer?

Dangelo Mann · Answer 1

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)