Differentiate y = 10x^3 - x^2 + 4x - 15 and then solve y' = 0.

Step-by-step solution

1. 1

   Apply the power rule term by term. For a polynomial each term is handled independently using $\left(x^{n}\right)^{\prime}=nx^{n-1}$ and the fact that constants differentiate to zero: $\left(10x^{3}\right)^{\prime}=30x^{2}$, $\left(-x^{2}\right)^{\prime}=-2x$, $(4x)^{\prime}=4$, $(-15)^{\prime}=0$.

2. 2

   Write down the derivative. Adding the four pieces, $y^{\prime}=30x^{2}-2x+4 .$

3. 3

   Set the derivative to zero and simplify. Solving $y^{\prime}=0$ means solving $30x^{2}-2x+4=0$. Every coefficient is even, so divide by $2$: $15x^{2}-x+2=0 .$

4. 4

   Test the discriminant instead of forcing a root. With $A=15$, $B=-1$, $C=2$, $D=B^{2}-4AC=(-1)^{2}-4(15)(2)=1-120=-119<0 .$ A negative discriminant means the quadratic has no real roots.

5. 5

   Interpret the result geometrically. Since $y^{\prime}=30x^{2}-2x+4$ never vanishes and its leading coefficient is positive, $y^{\prime}>0$ for every real $x$. So the cubic is strictly increasing and has no horizontal tangent and no local extremum anywhere.