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How do you show that the curve #6x^3+5x-3# has no tangent line with slope 4?

Answer 1

Aurora Alvarado

Find the derivative. Set it equal to #4#. Explain why there is no (real) solution.

For #y=6x^3+5x-3#, the slope of the tangent line is given by

#y' = 18x^2+5#.

In order to have a tangent line with sloe #4# there would have to be an #x# such that

#18x^2+5=4#.

But that is true only if #x^2 = -1/18#.

For real numbers #x#, we cannot have #x^2# is a negative number.

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Answer 2

Charles Arocha

To show that the curve 6x^3+5x-3 has no tangent line with slope 4, we can find the derivative of the curve and set it equal to 4. If there are no solutions to this equation, it means there is no tangent line with slope 4. The derivative of the curve is 18x^2 + 5. Setting this equal to 4, we get 18x^2 + 5 = 4. Solving this equation, we find that there are no real solutions for x. Therefore, the curve 6x^3+5x-3 has no tangent line with slope 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.