Question

Find the least value of 1/ 4−cos3​

Answer 1

Answer:

f(x)=cos3x−15cosx+8 where xϵ[

3

π

,

2

3π

]

f

′

(x)=−3sin3x+15sinx=0

⇒sin3x=5sinx

⇒3sinx−4sin

3

x=5sinx

⇒−4sin

3

x=2sinx

⇒2sinx+4sin

3

x=0

⇒2sinx[1+2sin

2

x]=0

Now, sinx=0 and sin

2

x=

2

−1

Not possible

x=π is the only choice because

xϵ[

2

π

,

2

3π

]

f

′′

(x)=−9cos3x+15cosx

f

′′

(π)=−6<0, therefore x=π is the point of maxima.

f(

2

π

)=cos

2

3π

−15cos

2

π

+8=8

f(π)=cos3π−15cosπ+8=−1+15+8=22

f(

2

3π

)=cos

2

9π

−15cos

2

3π

+8=8

x=π,f(x)=22 is the greatest value

x=

2

π

and x=

2

3π

,f(x)=8 is the least value.

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