SOLUTION: (a) use the method of first principal of differentiation to find the derivative of y=f(x)= 9x-x^3. hence find the equation of the tangent line to the curve y= f(x)at x= 1
(b
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Question 1118506: (a) use the method of first principal of differentiation to find the derivative of y=f(x)= 9x-x^3. hence find the equation of the tangent line to the curve y= f(x)at x= 1
(b) find another point (x,y)on the curve y=f(x)=9x-x^3 where the tangent is parallel to the tangent in (a)above
Found 2 solutions by Boreal, solver91311:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The derivative is 9-3x^2
where x=1, the slope is 9-3(1)=6
where x=1, f(x)=8
equation of a line with slope 6 and (1, 8) use point slope formula y-y1=m(x-x1) where m is slope and (x1, y1) point
y-8=6(x-1)
y-8=6x-6 or y=6x+2
Parallel lines have the same slope, so need another place where slope is 6
9-3x^2=6
3x^2=3
x=+/-1
when x is -1, f(x)=-8
line tangent is y-(-8)=6(x+1), or y+8=6x+6 or y=6x-2
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ -\ f(x)}{\delta x})
\,-\,(x\,+\,\delta x)^3\]\ -\ \[9x\,-\,x^3\]}{\delta x})
\ =\ 9(1)\ -\ 1^3\ =\ 8)
Slope of tangent at )
^2\ =\ 6)
Point-Slope form
)
Part B:
The slopes of parallel lines are equal. Hence, you need to find another value of 
such that 
. We already know that one of the roots of this equation is 
. Find the other. Then find the value of )
. Then derive an equation of the line that passes through the point \))
with a slope of 
John
My calculator said it, I believe it, that settles it
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