Question

Find the point on the curve \[ y = x^3 - 11x + 5 \] at which the tangent is \[ y = x - 11. \]

Hint:

Always verify the obtained points on both the curve and tangent line, as some extraneous solutions may appear.

Answer 1

Step 1: Slope of given tangent.
Equation of tangent line: \[ y = x - 11 $\Rightarrow$ \text{slope} = 1. \]

Step 2: Differentiate the curve.
Curve: \[ y = x^3 - 11x + 5 \] \[ \frac{dy}{dx} = 3x^2 - 11. \]

Step 3: Equating slope.
At tangent point, slope $= 1$: \[ 3x^2 - 11 = 1 \] \[ 3x^2 = 12 $\Rightarrow$ x^2 = 4 $\Rightarrow$ x = \pm 2. \]

Step 4: Find corresponding $y$ values.
For $x=2$: \[ y = 2^3 - 11(2) + 5 = 8 - 22 + 5 = -9. \] For $x=-2$: \[ y = (-2)^3 - 11(-2) + 5 = -8 + 22 + 5 = 19. \]

Step 5: Check if points lie on tangent line.
Tangent: $y = x - 11$. For $(2,-9)$: RHS = $2-11=-9$, ✅ lies on line. For $(-2,19)$: RHS = $-2-11=-13 \neq 19$, ❌ not valid.

Final Answer: \[ \boxed{(2,-9)} \]