Question

f(x) = 10/cos-1(x). Use calculus to determine:

a) all critical values

b) any local extrema

c) any absolute extrema

d) the intervals where f is increasing/decreasing

e) any points of inflection rounded to the thousandths place

f) intervals where f is concave up/down

No interval was specified so I assumed 0<x<2pi

also I didn't get any concavity not sure if I'm right or not.

I got x= pi as the critical value as well as the relative and absolute minimum extrema.

would greatly appreciate some help

Answer 1

Interval must be taken as

as arccos(x) or inverse of cos(x) is defined in domain [-1,1]

with range [0, pi]

as we can not have 0 in denomiantor, therefore x can not be equal to 1, this is the reason .

as arccos(x) and square root of 1-x^2 , both are always positive. therefore f'(x) > 0,

hence no critical root, and f(x) is increasing function in tis domain.

with lowest value when x = -1, that is f(x) = 10/pi = 3.1831.

On solving from calculator :

f''(x) = 0 at x ~=-0.6580

The point of inflection is x = -0.6580.

There is no critical value as f'(x) can not be zero

hence, there is no local extrema

Absolute extrema : minimum when x = -1, that is y = 10/pi maximum when x approaches 1, y approaches infinity.

f is increasing function in its domain that is

Point of inflection when f''(x) = 0 (in the neighborhood f''(x) changes sign too), or concavity changes. at x = -0.6580

function is concave down function is cocave up

Graph: