ANSWER
a.) The Gradient at the point C is 16
b.) Possible coordinate of P is (2, -13)
EXPLANATION
Given the equation of the curve C:
[tex]y=x^3-11x+1[/tex]The Gradient at the point on C where x = 3
[tex]\begin{gathered} y=x^3\text{ - 11x + 1} \\ \text{Gradient = }\frac{dy}{dx}=3x^2\text{ - 11} \\ \text{ }at\text{ x = 3,} \\ \text{Gradient = }\frac{dy}{dx}=3(3)^2\text{ - 11} \\ \text{Gradient = }\frac{dy}{dx}=27\text{ - 11} \\ \text{Gradient = }\frac{dy}{dx}=16 \end{gathered}[/tex]Determine the value of x when the point P lies on C with 1 being the gradient at that point
[tex]\begin{gathered} \text{Gradient = }\frac{dy}{dx}=3x^2\text{ - 11} \\ \text{1 }=3x^2\text{ - 11} \\ \text{3x}^2\text{ = 1+11} \\ 3x^2\text{ = 12} \\ x^2\text{ = }\frac{12}{3} \\ x^2\text{ = 4} \\ x\text{ = }\sqrt[]{4} \\ x\text{ = 2} \end{gathered}[/tex]Determine the value of y
[tex]\begin{gathered} y=x^3\text{ - 11x + 1} \\ y=(2)^3\text{ - 11(2) + 1} \\ y\text{ = 8 - 22 + 1} \\ y\text{ = -13} \end{gathered}[/tex]Hence, the possible coordinate of P is (2, -13)
