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Produce graphs of f that reveal all the important aspects of the curve. Then use calculus to find the following. (Enter your answers using interval notation. Round your answers to two decimal places.)
f(x) = 3 sin x + cot x, -[tex] - \pi \leq x \leq \pi [/tex]

a) find interval of decrease
b)find interval of increase
c)Find inflection points of the function and their y values
d) find interval where the function is concave up; where f is concave down

Respuesta :

syed514
a) Intervals of increase is where the derivative is positive
b)  Intervals of decrease is where the derivative is negative.  

c) Inflection points of the function are where the graph changes concavity that is the point where the second derivative is zero 

d)
Concave up- Second derivative positive 
Concave down- second derivative negative 

f(x) = 4x^4 − 32x^3 + 89x^2 − 95x + 31 
f '(x) = 16x^3 - 96x^2 + 178x - 95 
f "(x) = 48x^2 - 192x + 178 

By rational root theorem the f '(x) has one rational root and factors to: 
f '(x) = (2x - 5)*(8x^2 - 28x + 19) 

Using the quadratic formula to find it's two irrational real roots. 
The f "(x) = 48x^2 - 192x + 178 only has irrational real roots, use quadratic formula which will be the inflection points as well.

Preciousorekha1

Answer:

Step-by-step explanation:

given that the function is given from the question as;

f(x) = 3 sin x + cot x, where -π ≤ x ≤ π

solving;

given f(x) = 3 sin x + cot x

f(x) = 3 cos x - cos²x = 0

which is 3 cos x - 1/sin²x = 0

3 cos x = 1/sin²x

∴ 3 cosx.sin²x = 1

simplifying gives, 3cosx(1-cos²x) = 1

expanding gives 3cos³x - 3cosx + 1 = 0

given x = -1.165, -0.734, 0.734 , 1.165

here x is critical at 0

for region 1; (-Ï€, -1.165)

for region 2; (-1.165, -0.734) ⇒ increases

for region 3; (-0.734, 0) ⇒ decreases

for region 4; (0, 0.734) ⇒ decreases

for region 5; (0.734, 1.165) ⇒ increases

for region 6; (1.165,π) ⇒ decreases

(a). interval of decrease becomes : (Ï€,-1.165)U(-0.734,0)U(0,0.734)U(1.165,Ï€)

(b). interval of increase becomes : (-1.165,-0.734)U(0.734,1.165)

(c). to solve for the inflection points, we input the equation

f(x) = sin x (2 cotx cos³x -3)

inputting we have that

x = -0.922 and 0.922

x = 0

for region 1; (-π, -0.922) ⇒ concave up

for region 2; (-0.922, 0) ⇒ concave down

for region 3; (0,0.922) ⇒ concave up

for region 4; (0.922,π) ⇒ concave down

(d). For concave up; (-Ï€,-0.922)U(0,0.922)

     For concave down; (-0.922,0)U(0.992, π)

cheers i hope this helps

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