f(x)= sin (squared) (x/2) defined on the interval [-5.7831852, 1.3707963] find maximum and minimum points and inflection points.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Note that the sine function is always defined on the interval [-1,1].

You need to calculate the first derivative of the function to find the extreme point.

`f'(x) = (sin sqrt(x/2))'`

Use chain rule=> `f'(x) = (cos sqrt(x/2))(sqrt(x/2))'*(x/2)'`

`f'(x) = (cos sqrt(x/2))(1/2sqrt(x/2))*(1/2)`

`f'(x) = [cos sqrt(x/2)]/[4sqrt(x/2)]`

Put`f'(x)=0 =gt [cos sqrt(x/2)]/[4sqrt(x/2)] = 0 =gt cos sqrt(x/2)]=0`

`sqrt(x/2) = pi/2 + npi`

`x/2 = pi^2/4 + n*pi^2 + n^2*pi^2`

`x = pi^2/2 + 2*n*pi^2 + 2*n^2*pi^2`

The extreme point is `x = pi^2/2 + 2*n*pi^2 + 2*n^2*pi^2`

You need to find the second derivative of f(x) to find the inflection point.

`f"(x) = (cos sqrt(x/2)/4sqrt(x/2))' `

`f"(x) = {(cos sqrt(x/2))'*4sqrt(x/2) - cos sqrt(x/2)*(4sqrt(x/2))'}/(8x)`  (quotient rule)

`f"(x) = {-sin sqrt(x/2)*(1/4sqrt(x/2))*4sqrt(x/2) - [cos sqrt(x/2)]*4/4sqrt(x/2)}/(8x)`

`f"(x) = {-sin sqrt(x/2) - [cos sqrt(x/2)]/sqrt(x/2)}/(8x)`

Put `f"(x) = 0`

`{-sin sqrt(x/2) - [cos sqrt(x/2)]/sqrt(x/2)}/(8x) = 0`

`-sin sqrt(x/2) - [cos sqrt(x/2)]/sqrt(x/2) = 0`

`-sin sqrt(x/2)*(sqrt(x/2)) - cos sqrt(x/2) = 0`

Divide by `cos sqrt(x/2):`

`- (sqrt(x/2))*tan sqrt(x/2) - 1 = 0`

`tan sqrt(x/2) = -1/(sqrt(x/2))`

`sqrt(x/2) = arctan (-1/(sqrt(x/2)))`

`x/2 = {arctan (-1/(sqrt(x/2)))}^2`

`x = 2{arctan (-1/(sqrt(x/2)))}^2`

ANSWER: The extreme point is `x = pi^2/2 + 2*n*pi^2 + 2*n^2*pi^2`  and the inflection point is `x = 2{arctan (-1/(sqrt(x/2)))}^2`

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