Can you solve this calculus question

f(x) = x^2 - mx - 6 for x < 2, and f(x) = x^2 + 2 for x >= 2.

a) lim x -> 2^- of f(x) = lim x -> 2^+ of f(x)
(2)^2 - m(2) - 6 = (2)^2 + 2
-2m - 2 = 6
m = -4 makes f continuous everywhere.

b) For this value of m,
f(x) = x^2 + 4x - 6 for x < 2, and f(x) = x^2 + 2 for x >= 2.
To find out if f is differentiable at x = 2, we need to compare
lim x -> 2^- of [f(x) - f(2)]/(x - 2) and lim x -> 2^+ of [f(x) - f(2)]/(x - 2).
lim x -> 2^- of [f(x) - f(2)]/(x - 2) = 2(2) + 4 = 8.
lim x -> 2^+ of [f(x) - f(2)]/(x - 2) = 2(2) = 4.
Since 8 is unequal to 4, f is not differentiable at x = 2.

Lord bless you today!

f(x)=x² - mx - 6, x<2
f(x)=x² + 2, x≥2

Lim (x-->2-) f(x) = Lim (x-->2-) x² - mx - 6 = 2² - 2m - 6 = - 2m - 2
Lim (x-->2+) f(x) = Lim (x-->2-) x² + 2 = 2² + 2 = 6

- 2m - 2=6 => m =-4

Hence
f(x)=x² + 4x - 6, x<2
f(x)=x² + 2, x≥2

Case x<2
d(f(x))/dx = (x² + 4x - 6)` = 2x + 4
f`(2)=8

Case x≥2
d(f(x))/dx = (x² + 2)` = 2x
f`(2)=4

8≠4
Hence
f(x) does not differentiable