Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 39156 by rahul 19 last updated on 03/Jul/18

Answered by MJS last updated on 03/Jul/18

f(x)=c_3 x^3 +c_2 x^2 +c_1 x+c_0 f′(x)=3c_3 x^2 +2c_2 x+c_1 we have 1. f′(−1)=0 2. f′(1)=0 3. f(−1)=10 4./f(3)=−22 1. 3c_3 −2c_2 +c_1 =0 2. 3c_3 +2c_2 +c_1 =0 3. −c_3 +c_2 −c_1 +c_0 =10 4. 27c_3 +9c_2 +3c_1 +c_0 =−22 1. c_1 =2c_2 −3c_3 2. 3c_3 +2c_2 +2c_2 −3c_3 =0 ⇒ c_2 =0; c_1 =−3c_3 3. 2c_3 +c_0 =10 ⇒ c_0 =10−2c_3 4. c_3 =−2 ⇒ c_0 =14 ⇒ c_1 =6 f(x)=−2x^3 +6x+14 horizontal tangents pass through (((−1)),((f(−1)=10)) ) and ((1),((f(1)=18)) ) distance=8; (1/4) distance=2

f(x)=c3x3+c2x2+c1x+c0f′(x)=3c3x2+2c2x+c1wehave1.f′(−1)=02.f′(1)=03.f(−1)=104./f(3)=−221.3c3−2c2+c1=02.3c3+2c2+c1=03.−c3+c2−c1+c0=104.27c3+9c2+3c1+c0=−221.c1=2c2−3c32.3c3+2c2+2c2−3c3=0⇒c2=0;c1=−3c33.2c3+c0=10⇒c0=10−2c34.c3=−2⇒c0=14⇒c1=6f(x)=−2x3+6x+14horizontaltangentspassthrough(−1f(−1)=10)and(1f(1)=18)distance=8;14distance=2

Commented by rahul 19 last updated on 03/Jul/18

Thank you sir :)

Thankyousir:)

Terms of Service

Privacy Policy

Contact: info@tinkutara.com