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The gradient of the tangent to the curve y= ax^3 + bx at the point ( 1, 1) is -5. Calculate the values of a and bAsked by kudu
The gradient of the tangent to the curve y= ax^3 + bx at the point ( 1, 1) is -5. Calculate the values of a and b
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Answered by
MathMate
We need to find a and b, so we will look for two equations.
The first one is simply the point (1,1), which tells us that 1=y(1), or
1=a(1^3)+b(1)=a+b, or
a+b=1............(1)
The second one is the fact that the gradient (dy/dx) at (1,1)=-5.
Since dy/dx = 3ax²+b, we have
-5 = 3a(1^2)+b = 3a+b
or
3a+b = -5...........(2)
Solve for a and b given the system of equations:
a+b=1............(1)
3a+b = -5...........(2)
You can solve by elimination.
The first one is simply the point (1,1), which tells us that 1=y(1), or
1=a(1^3)+b(1)=a+b, or
a+b=1............(1)
The second one is the fact that the gradient (dy/dx) at (1,1)=-5.
Since dy/dx = 3ax²+b, we have
-5 = 3a(1^2)+b = 3a+b
or
3a+b = -5...........(2)
Solve for a and b given the system of equations:
a+b=1............(1)
3a+b = -5...........(2)
You can solve by elimination.
Answered by
Anonymous
a=-3 ,b=4
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