Question:
Show that the curve y=6x^3+5x-3 has no tangent line with slope 4.
Known:
Well of course we know that the y' is 18x^2+5, so since x=sqrt(-5/18), would that prove that the curve has no tangent line because x has an imaginary number?
or
Would you set the derivative equal to 4? ---- 4=18x^2+5
This would still give you an imaginary number, x=sqrt(-1/18). Would this be the proof?
or
Am I just totally off here? I am not to sure. I kind of think that the second thing I did would show why there is no tangent lin on the curve, but I am not to sure.
Would someone mind helping me out on this problem. It is simple I know, but I think I may be thinking to much in it or something. I appreciate the help in advance
Show that the curve y=6x^3+5x-3 has no tangent line with slope 4.
Known:
Well of course we know that the y' is 18x^2+5, so since x=sqrt(-5/18), would that prove that the curve has no tangent line because x has an imaginary number?
or
Would you set the derivative equal to 4? ---- 4=18x^2+5
This would still give you an imaginary number, x=sqrt(-1/18). Would this be the proof?
or
Am I just totally off here? I am not to sure. I kind of think that the second thing I did would show why there is no tangent lin on the curve, but I am not to sure.
Would someone mind helping me out on this problem. It is simple I know, but I think I may be thinking to much in it or something. I appreciate the help in advance
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