Question

Solve the following : 

f(x) = –x, for – 2 ≤ x < 0
= 2x, for 0 ≤ x < 2
= `(18 - x)/(4)`, for 2 < x ≤ 7
g(x) = 6 – 3x, for 0 ≤ x < 2
= `(2x - 4)/(3)`, for 2 < x ≤ 7
Let u (x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v' (1) and w'(1). If it doesn't exist, then explain why?

Answer 1

u(x) = f[g(x)]
∴ `u'(x) = "d"/"dx"{f[g(x)}`

= `f'[g(x)]."d"/"dx"[g(x)]`
= f'[gx)] x g'(x)
∴ u'(1) = f'[g(1)] x g'(1)
= f'(3) x g'(1)                  ...(1)
                       ...[∵ g(x) = 6 – 3x, 0 ≤ x ≤ 2]
Now, f(x) = `(18 - x)/(4)`, for 2 < x ≤ 7
and g(x) = 6 – 3x, for 0 < x ≤ 2
∴ f'(x) = `(1)/(4)(0 - 1) = -(1)/(4)`, for 2 < x ≤ 7
and g'(x) = 0 – 3(1) = – 3, for 0 < x ≤ 2
∴ `f'(3) = -(1)/(4) and g'(1)` = – 3
∴ from (1),
u'(1) = `-(1)/(4)(-3) = (3)/(4)`
Now, v(x) = g[f(x)]
∴ v'(x) = `"d"/"dx"{g[f(x)]}`

= `g'[f(x)]."d"/"dx"[f(x)]`
= g'[f(x)] x f'(x)
∴ v'(1) = g'[f(1)] x f'(x)
= g'(2) x f'(1)                  ...(2)
                       ...[∵ f(x) = 2x, 0 ≤ x ≤ 2]
Now, g(x) = 6 – 3x, for 0 ≤ x ≤ 2
= `(2x - 4)/(3)`, for 2 < x ≤ 7
∴ g"(x) = 0 – 3 x 1 = – 3, for 0 ≤ x ≤ 2
and g'(x) = `(1)/(3)(2 xx 1 - 0) = (2)/(3)`, for 2 < x ≤ 7
∴ Lg'(2) ≠ Rg'(2)
∴ g'(2) does not exist
∴ from (2),
v'(1) does not exist
Also, w(x) = g[g(x)]
∴ w'(x) = `"d"/"dx"{g[g(x)]}`

= `g'[g(x)]."d"/"dx"[g(x)]`
= g'[g(x)] x g'(x)
∴ w'(1) = g'[g(1)] x g'(x)
= g'(3) x g'(1)                ...(3)
                    ...[∵ g(x) = 6 – 3x, 0 ≤ x ≤ 2]
Now, g(x) = 6 –3x, for 0 ≤ x ≤ 2
= `(2x - 4)/(3)`, for 2 < x ≤ 7
∴ g'(x) = 0 – 3 x 1 = – 3, for 0 ≤ x ≤ 2
and g'(x) = `(1)/(3)(2 xx 1 - 0) = (2)/(3)`, for 2 ≤ x ≤ 7
∴ g(3) = `(2)/(3) and g'(1)` = – 3
∴ from (3),
w'(1) = `(2)/(3)(-3)` = – 2.
Hence, u'(1) = `(3)/(4)`, v'(1) does not exist and w'(1) = – 2.