Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 73

Prove that the curve $y = 6x^3 + 5x - 3$ has no tangent line that has a slope of 4.

$\begin{equation} \begin{aligned} \text{Given: } y &= 6x^3+5x-3\\ y' &= m_T = 4 \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} y &= 6x^3 + 5x -3\\ y' &= m_T = 6 \frac{d}{dx} (x^3) + 5 \frac{d}{dx} (x) - \frac{d}{dx} (3) && \text{Derive each terms}\\ \\ &= (6)(3x^2)+(5)(1) - 0 && \text{Simplify the equation}\\ \\ m_T &= 17x^2 + 5 && \text{ Substitute the given value of the slope } (m_T)\\ \\ 18x^2+5 &= 4 && \text{Add -5 to each sides}\\ \\ 18x^2 &= 4- 5 && \text{Combine like terms and divide both sides by 18}\\ \\ \frac{18x^2}{18} &= \frac{-1}{18} && \text{Take the square root of both sides}\\ \\ \sqrt{x^2} &= \sqrt{\frac{-1}{18}}\\ x &= \pm \sqrt{\frac{-1}{18}} && \text{The curve has no tangent line because the values of } x \text{ are invalid} \end{aligned} \end{equation}$