Find the point on the curve $y = x^3 - 11 x + 5$ at which the target is $y = x -11.$

Question

$(A)\; 2, 9$
$(B)\; 2,-9$
$(C)\; -2, 9 $
$(D)\; -2,-9$

Answer 1

Toolbox:

• If $y = f(x)$, then $(\frac{dy}{dx})P =$ slope of the tangent to $y = f(x)$ at that point $P$.
• Equation of the line with given two points is $\frac{x-x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1}$

STEP 1:

Given : $y = x^3-11x+5$

Differentiate w.r.t $x$ we get,

$\frac{dy}{dx} = 3x^2-11 ------(1)$

The equation of the given tangent is $y = x-11$

STEP 2:

Hence the slope of the tangent is 1

Now Equating this slope to equ (1)

$\begin{align*} 3x^2 -11 & = 1 \\ 3x^2 & = 12 \\ x^2 & = 4 \\ x & = \pm 2 \end{align*}$

STEP 3:

When $x = 2, y = -9$

When $x = -2, y = -13$

Hence the point is $(2, -9)$