Calc 1 question

Question

what are the concaves up and down and the inflection points step by step for ln(2+sin(x))

Answer 1

f(x) = ln(2 + sinx), 0 < x < 2π

f'(x) = cosx / (2 + sinx)

f"(x) = [(2+sinx)(-sinx) - cosx(cosx)] / (2 + sinx)² = [-2sinx - (sin²x + cos²x)] / (2+sinx)²

f"(x) = [-2sinx - 1] / (2+sinx)² = 0

sinx = -1/2 x = 7π/6, 11π/6

If 0 < x < 7π/6, then f"(x) < 0. So f is concave down.

If 7π/6 < x < 11π/6, then f"(x) > 0. So, f is concave up.

If 11π/6 < x < 2π, then f"(x) < 0. So, f is concave down.

Answer 2

find the first and second derivatives to find local extrema and inflection points

y = ln ( 2 + sin(x) )

Recall d/dx ln(u) = (1/u)*u'.

y' = cos(x) / ( 2 + sin (x) )

Set the fraction, and by limitation, the numerator equal to zero.

cos(x) = 0, therefore

x=π/2 ± 2πn (maximums)

x=3π/2± 2πn (minimums)

Recall quotient rule d/dx (u/v) = (u'v - v'u) / v²

y'' = (-sinx(2+sinx) - cos²x)/(2+sinx)²

Set the fraction, and by limitation, the numerator equal to zero.

0 = -sinx(2+sinx) - cos²x

cos²x = -sin(2+sinx)

Distribute the -sinx

cos²x = -2sin²x - 2sinx

cos²x + sin²x = -2sinx

Use the Trig/Pythag identity and divide both sides by 2

-1/2 = sinx

Recall special angles to determine inflection values of x

x = 7π/6 ± 2πn

x = -π/6 ± 2πn

Since the maximums π/2 ± 2πn lie between -π/6± 2πn and 7π/6 ± 2πn , those regions are concave down.

Since the minimums 3π/2 ± 2πn lie between 7π/6 ± 2πn and 11π/6 ± 2πn, those regions are concave up.