Question

5. Verity Roile's mean value theorem f(x) = x(x + 3)e x/2 (-3, 0]?​

Answer 1

Answer:

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Step-by-step explanation:

We have f(x)=x(x+3)e

We have f(x)=x(x+3)e −x/2

We have f(x)=x(x+3)e −x/2

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]=− 21 [x 2 −x−6]e −x/2

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]=− 21 [x 2 −x−6]e −x/2

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]=− 21 [x 2 −x−6]e −x/2 f (x) exist for every value of x in the interval[-3,0]

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]=− 21 [x 2 −x−6]e −x/2 f (x) exist for every value of x in the interval[-3,0]Hence, f(x) is differentiable and hence, continous in the interval [−3,0]

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]=− 21 [x 2 −x−6]e −x/2 f (x) exist for every value of x in the interval[-3,0]Hence, f(x) is differentiable and hence, continous in the interval [−3,0]Also, we have f(−3)=f(0)=0⇒All the three conditions of Rolle's theorem are satisfied.

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]=− 21 [x 2 −x−6]e −x/2 f (x) exist for every value of x in the interval[-3,0]Hence, f(x) is differentiable and hence, continous in the interval [−3,0]Also, we have f(−3)=f(0)=0⇒All the three conditions of Rolle's theorem are satisfied.So f (x)=0⇒ 21 (x 2−x−6)e −x/6

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]=− 21 [x 2 −x−6]e −x/2 f (x) exist for every value of x in the interval[-3,0]Hence, f(x) is differentiable and hence, continous in the interval [−3,0]Also, we have f(−3)=f(0)=0⇒All the three conditions of Rolle's theorem are satisfied.So f (x)=0⇒ 21 (x 2−x−6)e −x/6 =0⇒x 2−x−6=0⇒x=3,−2

We have f(x)=x(x+3)e −x/2 ∴f (x)=(2x+3)e −x/2 +(x 2+3x)e −x/2 (− 21 )=e −x/2 [2x+3− 21[x 2+3x]]=− 21 [x 2 −x−6]e −x/2 f (x) exist for every value of x in the interval[-3,0]Hence, f(x) is differentiable and hence, continous in the interval [−3,0]Also, we have f(−3)=f(0)=0⇒All the three conditions of Rolle's theorem are satisfied.So f (x)=0⇒ 21 (x 2−x−6)e −x/6 =0⇒x 2−x−6=0⇒x=3,−2Since, the value x=−2 lies in the open interval [−3,0] the Rolle's theorem is verified