Question

In: Math

f(x) = 10/cos-1(x). Use calculus to determine: a) all critical values b) any local extrema c)...

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

Expert Solution

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:

milcah answered 2 years ago

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