IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

naman

Find a formula for a function g(x) satisfying the following conditions

a] domain of g is (-

,+

)

b] range of g is [-2,8]

c] g(x) has a period "pie"

d] g(2) =3

titun

There can be many functions to satisfy the given conditions.

iitdreams

hi yaar ,its very easy.....[5sin2(x-2)] +3

iitdreams

i think this is the answer ur lookin for

puneet

hii

ya the correct answer is alrealy there ..

cheero

iitkgp_bipin

Since its domain is unbounded and range is bounded assume it to be sinusoidal function.

Function of the form Asin(x+k) + b has period 2pi.

But here the period is pi , so replace x by 2x.

g(x) = Asin(2x+k) + B

max{g(x)} = A + B = 8

min{g(x)} = -A + B = -2

these two give : A=5 and B=3

g(x) = 5sin(2x+k) + 3

now last condition :

g(2) = 3

5sin(4+k) + 3 = 3

sin(4+k) = 0

k = -4

Hence g(x) = 5sin(2x-4) + 3

Many other functions are also possible of the form Asin(2x+k)+Bcos(2x+k)+C.