First there's some correction in your solution
[tex]\frac{dy}{dx}=3x^2-11[/tex]
and you have put value x=3 in
[tex]\begin{gathered} \frac{dy}{dx}=3x^2 \\ \frac{dy}{dx}=3\times3^2 \\ \frac{dy}{dx}=27 \end{gathered}[/tex]
the correct calculation will be,
[tex]\begin{gathered} \frac{dy}{dx}=3x^2-11 \\ \frac{dy}{dx}=3\times3^2-11 \\ \frac{dy}{dx}=27-11 \\ \frac{dy}{dx}=16 \end{gathered}[/tex]
Now, answer of the (b) part is,
Given curve is,
[tex]y=x^3-11x+1[/tex]
and point P lies on C and the gradient at that point is 1.
We will take point P on curve C as
[tex](x_{1,}y_1)[/tex]
Put point P in curve C, we will get,
[tex]y_1=x_1^3-11x_1+1[/tex]
Now, we will differentiate
[tex]y_1[/tex]
we will get,
[tex]\frac{dy}{dx}=3x^2_1-11_{^{^{}}}[/tex]
Now, we have given that gradient at point P is 1,
[tex]\begin{gathered} 1=3x^2_1-11 \\ 3x^2_1=12 \\ x^2_1=4 \\ x_1=\pm2 \end{gathered}[/tex]
corresponding these points we will get y=
[tex]\pm14[/tex]
So, possible co-ordinate of P is
[tex](\pm2,\pm14)[/tex]
