## Calculus (3rd Edition)

Let f(x)= y. Slope of the tangent to the given curve at (x,y) is $\frac{dy}{dx}$= $3x^{2}-6x+1$ The slope is given to be 10. So $3x^{2}-6x+1$=10 ( As the slope of the tangent to the curve y= f(x) at the point ($x_{0},y_{0}$) is given by f'($x_{0}$).) Or $3x^{2}-6x-9=0$ This gives x= 3 or x= -1 Now, y= $x^{3}-3x^{2}+x+4$ So when x=3, y=$27-(9\times3)+3+4= 7$ When x= -1, y= -1-3-1+4= -1 Therefore, the required points are (3,7) and (-1,-1)