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Hi please can I get help finding the P coordinates as I’m stuck with the second half of the question

Hi please can I get help finding the P coordinates as I’m stuck with the second half-example-1

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6 votes

First there's some correction in your solution


(dy)/(dx)=3x^2-11

and you have put value x=3 in


\begin{gathered} (dy)/(dx)=3x^2 \\ (dy)/(dx)=3*3^2 \\ (dy)/(dx)=27 \end{gathered}

the correct calculation will be,


\begin{gathered} (dy)/(dx)=3x^2-11 \\ (dy)/(dx)=3*3^2-11 \\ (dy)/(dx)=27-11 \\ (dy)/(dx)=16 \end{gathered}

Now, answer of the (b) part is,

Given curve is,


y=x^3-11x+1

and point P lies on C and the gradient at that point is 1.

We will take point P on curve C as


(x_(1,)y_1)

Put point P in curve C, we will get,


y_1=x_1^3-11x_1+1

Now, we will differentiate


y_1

we will get,


(dy)/(dx)=3x^2_1-11_{^{^{}}}

Now, we have given that gradient at point P is 1,


\begin{gathered} 1=3x^2_1-11 \\ 3x^2_1=12 \\ x^2_1=4 \\ x_1=\pm2 \end{gathered}

corresponding these points we will get y=


\pm14

So, possible co-ordinate of P is


(\pm2,\pm14)

User Reavis
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