The equation of a curve is y=1/3x³-2x²-3x+5. (i)Determine the stationary points of the curve. (ii)For each point in a(i) above determine the nature of the points. (b)Find the...

      

The equation of the curve is y=1/3x³-2x²+3x+5.
(a)(i)Determine the stationary points of the curve.
(ii)For each point in (a)(i) above determine the nature of the points.
(b)Find the equation of the tangent to the curve at x=2.

  

Answers


Seline
(a)(i)
dy/dx=x²-4x+3
x²-4x+3=0
x²-3x-x+3=0
(x-1)(x-3)=0
x-1=0 or x-3=0
x=1 x=3
x=1 y=1/3-2+3+5=6,1/3 pt(1, 6,1/3)
x=3 y=9-18+9+5=5 pt(3, 5)

(ii)d²y/dx²=2x-4

x=1 d²y/dx²=-2 maximum(1, 6,1/3)

x=3 d²y/dx²=6-4=2 minimum(3,5)

(b)dy/dx=x²-4x+3
x=2 dy/dx=4-8+3=-1
y=8/3-8+6+5=5,2/3
pt(2, 5,2/3)

y-5,2/3 / x-2 = -1
3x+3y=23


Belan answered the question on February 25, 2019 at 14:56


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