The slope of the tangent line is the value of the derivative. The derivative is
f ′ ( x ) = 3 x 2 + 12. {\displaystyle f'(x)=3x^{2}+12.}
Solving 3 x 2 + 12 = 15 {\displaystyle 3x^{2}+12=15} , gives the solutions x = − 1 {\displaystyle x=-1} and x = 1 {\displaystyle x=1} .
(The algebra for this problem is given below)
3 x 2 + 12 = 15 ⇒ 3 x 2 = 3 ⇒ x 2 = 1 ⇒ x = 1 {\displaystyle 3x^{2}+12=15\Rightarrow 3x^{2}=3\Rightarrow x^{2}=1\Rightarrow x=1} or x = − 1 {\displaystyle x=-1} .
