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.)

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)

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it