Determine where the graph of the function f(x)=sin9x is concave up and concave down?

Question

On the interval 0 ≤ x ≤ 2π/9. Also find all inflection points of the function

 

Answer 1

The trigonometric function is f(x) = sin(9x) and the interval is [0, 2π/9].

Differentiate with respect to x.

f'(x) = 9cos(9x).

Differentiate with respect to x.

f''(x) = -81sin(9x).

TEST FOR CONCAVITY :

If f " (x) > 0 (Positive) for all x in (a, b), then f(x) is concave upward on (a, b).

If f " (x) < 0 (Negative) for all x in (a, b), then f(x) is concave downward on (a, b).

To apply theorem, locate the x - values at which f " (x) = 0 or f " (x) does not exist.

f " (x) = 0 ⇒sin(9x) = 0.

Let 9x = t ⇒sin(t) = 0⇒sin(t) = sin(0^o).

General solution: θ = 180^on + (-1)ⁿα, where n is an integer.

α = 0^o⇒ t = 180^on + (-1)ⁿ(0) ⇒ t = 180^on ⇒ 9x = 180^on ⇒ x = 20^on, where n is an integer.

The solutions are . . . . . , -40^o, -20^o, 0, 20^o, 40^o, . . . . . .

The solutions in the interval [0, 2π/9] are x = 0^o, x = 20^o and x = 40^o.

Test intervals  x - Value            Polynomial or f " (x) value           Conclusion

(-40^o, -20^o)      x = -30          f''(-30) = -81sin(-270) = -81< 0   Concave downward

(-20^o, 0^o)          x = -10          f''(-10) = -81sin(-90) = 81> 0      Concave upward.

(0^o, 20^o)           x = 10           f''(10) = -81sin(90) = -81< 0        Concave downward

(20^o, 40^o)         x = 30           f''(30) = -81sin(270) = 81> 0       Concave upward.

 

Points of Inflection : If (c, f(c)) is a point of inflection of the graph of f(x), then either f " (c) = 0 or f " (x) does not exist.

The solutions in the interval [0, 2π/9] are x = 0^o, x = 20^o and x = 40^o.

If x = 0^o ⇒ then, f(0) = sin(0) = 0.

If x = 20^o ⇒ then, f(0) = sin(180) = 0.

If x = 40^o ⇒ then, f(0) = sin(360) = 0.

The Inflection points are (0^o, 0), (20^o, 0) and (40^o, 0).