(v) First Derivative Analysis.
Start with f(x) = (x^2 + 4x - 15) + 18/(x + 4) [for x ≠ -4, 1].
So, f '(x) = (2x + 4) - 18/(x + 4)^2.
Plainly f ' is undefined at x = -4, 1.
(For x = 1, we need not place it on a number line, because it is a removable discontinuity.)
Next, f '(x) = 0
==> (2x + 4) - 18/(x + 4)^2 = 0
==> x + 2 = 9/(x + 4)^2
==> (x + 2)(x + 4)^2 - 9 = 0
==> x^3 + 10x^2 + 32x + 23 = 0
==> (x + 1) (x^2 + 9x + 23) = 0
==> x = -1 (other two roots have nonzero imaginary parts).
This (and x = -4) breaks the the number line into three parts.
If x < -4, then f '(x) < 0 (try x = -1000).
If -4 < x < -1, then f '(x) < 0 (try x = -2)
If x > -1 (excluding x = 1, the hole), then f '(x) > 0 (try x = 0).
So, we have a local minimum at x = -1 (with value f(-1) = -12).
Otherwise, f is increasing for x in (-1, 1) U (1, ∞), and
f is decreasing for (-∞, -4) U (-4, -1).
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I hope this helps!