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}
$