Answer
No tangent line exists.
Work Step by Step
Using both Addition and Power rules, we can find $f'(x)$ as follows: $$f(x) = x^{5} + 3x^{3} + 5x$$ $$f'(x) = 5x^{5-1} + 9x^{3-1} + 5x^{1-0}$$ $$f'(x) = 5x^{4} + 9x^{2} + 5$$ Since the exercise is asking us to disprove the existence of $x$ such that $f'(x) = 3$, we can do the following: $$f'(x) = 3 = 5x^{4} + 9x^{2} + 5$$ $$0 = 5x^{4} + 9x^{2} + 5 - 3$$ $$0 = 5x^{4} + 9x^{2} + 2$$ For simplicity, let us make $z = x^{2}$ so we can rewrite the equation above as follows: $$0 = 5z^{2} + 9z + 2$$ which is now a quadratic equation that can be solved using the quadratic formula: $$z = \frac{-b +/- [b^{2} - 4(a)(c)]^{\frac{1}{2}}}{2a}$$ $$z = \frac{-9 +/- [9^{2} - 4(5)(2)]^{\frac{1}{2}}}{2(5)}$$ $$z = \frac{-9 +/- [81 - 40]^{\frac{1}{2}}}{10}$$ $$z = \frac{-9 +/- \sqrt 41}{10}$$ Substituting $z=x^{2}$, we get: $$x^{2} = \frac{-9 +/- \sqrt 41}{10}$$ Before we can attempt to solve for $x$, however, we need to pay attention to the equation's numerator. Since: $$9^{2} > 7^{2} > 41 > 6^{2}$$ $$9>7>\sqrt 41 > 6$$ we can now plainly see that, for all possible solutions of $x$, the numerator will always be negative. Therefore, there are no Real number solutions to this exercise, proving that no tangential line with slope $m = 3$ exists for $f(x) = x^{5} + 3x^{3} + 5x$
