The equation of the given curve is y = x3 – 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,

dy

dx

= 3x2 - 11
Then, we have :
3x2 - 11
⇒ 3x2 = 12
⇒ x2 = 4
⇒ x = ± 2
When x = 2, y = (2)3 − 11 (2) + 5 = 8 − 22 + 5 = −9.
When x = −2, y = (−2)3 − 11 (−2) + 5 = −8 + 22 + 5 = 19.
Hence, the required points are (2, −9) and (−2, 19).
OR
Consider y = √x , Let x = 49 andΔx = 0.5
Then,
Δy = √x+Δx−√x
= √49.5−√49
= √49.5−7
⇒ √49.5 = 7 + Δy
Now, dy is approximately equal to Δy and is given by,
dy = (

dy

dx

) Δx
=

1

22√x

(05) [Since y = √x]
=

1

2√49

(0.5)
=

1

14

(0.5)
= 0.035
Hence the approximate value of √49.5 is 7 + 0.035 = 7.035