The equation of the given curve is y = ${\mathit{x}}^3$ – 11x + 5
The equation of the tangent to the given curve is given as y = x − 11 (which is of the form y = mx + c).
∴ Slope of the tangent = 1
Now, the slope of the tangent to the given curve at the point (x, y) is given by,
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $3{\mathit{x}}^2$ - 11
Then, we have :
$3{\mathit{x}}^2$ - 11
⇒ $3{\mathit{x}}^2$ = 12
⇒ ${\mathit{x}}^2$ = 4
⇒ x = ± 2
When x = 2, y = ${\left(2\right)}^3$ − 11 (2) + 5 = 8 − 22 + 5 = −9.
When x = −2, y = ${\left(-2\right)}^3$ − 11 (−2) + 5 = −8 + 22 + 5 = 19.
Hence, the required points are (2, −9) and (−2, 19).
OR
Consider y = $\sqrt{\mathit{x}}$ , Let x = 49 andΔx = 0.5
Then,
Δy = $\sqrt{\mathit{x}+\mathit{\Delta }\mathit{x}}-\sqrt{\mathit{x}}$
= $\sqrt{49.5}-\sqrt{49}$
= $\sqrt{49.5}-7$
⇒ $\sqrt{49.5}$ = 7 + Δy
Now, dy is approximately equal to Δy and is given by,
dy = $\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)$ Δx
= $\frac{1}{22\sqrt{\mathit{x}}}$ (05) [Since y = $\sqrt{\mathit{x}}$]
= $\frac{1}{2\sqrt{49}}$ (0.5)
= $\frac{1}{14}$ (0.5)
= 0.035
Hence the approximate value of $\sqrt{49.5}$ is 7 + 0.035 = 7.035