Given that y= x^3 – 4x^2 +5x - 2 (x is not times)?
Question: 1)find (dy)÷(dx)
the point P is on the curve and its x co-ordinates is 3
2)calculate the y co-ordinates of P.
3)calculate the gradient of P
4)find the equation of the tangent at P
5)find the equation of the average at P
6)find the values of x for which the curve has a incline of 5
Answers:
1) dy/dx = 3x^2 -8x +5
2) y = 27 - 36 + 15 - 2
y = 4
P = (3, 4)
3) dy/dx = 27 -24 + 5
dy/dx = 8
4) y = mx + c
4 = 8 (3) + c
c = -20
so y = 8x - 20
5) m = -1/8
4 = -1/8 (3) + c
c = 4 3/8
so 8y = 35 - x
6) dy/dx = 3x^2 -8x +5
5 = 3x^2 -8x +5
8x = 3x^2
either x = 0 or 8/3
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y= x^3 – 4x^2 +5x - 2
y = x^2(x-4) + 1(5x - 2)
y = (x^2+1)(x-4)(5x-2)
y(3) = 3^3 –4*3^2 +5*3 –2 =4; P=[3, 4];
y’(x) =3x^2 –8x +5; y’(3) =k(P)= 3*3^2 –8*3 +5 =8;
tangent y=k*x+b containing P; or; 4=8*3 +b, hence b=-20; thus y=8*x-20;
normal y=(-1/k)*x+b containing P; or’ 4 =-0.125*3 +b; hence b=4.375; thus y=-0.125*x +4.375; or; 8y+x=35;
y’(x)=5= 3x^2-8x+5; or; x(3x-8)=0; hence x1=0, x2=8/3;
y = x^3 - 4x^2 + 5x -2
1) dy/dx = 3x^2 - 8x + 5
2) x = 3 => y = 3^3 - 4(3^2) + 5(3) - 2 = 4
3) x = 3 => dy/dx = 3(3^2) - 8(3) + 5 = 8
4) (y - 4) = 8(x - 3) => y = 8x - 20
5) slant of normal = -1/8
(y - 4) = -(x - 3)/8 => y = -x/8 + 35/8
6) dy/dx = 5 = 3x^2 - 8x + 5
=> 3x^2 - 8x = 0
=> x(3x - 8) = 0
=> x = 8/3 or x = 0
y= x^3 – 4x^2 +5x - 2
y = x^2(x-4) + 1(5x - 2)
y = (x^2+1)(x-4)(5x-2)
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