1. a. No, g(x) is not a function because f(x) is not invertible. This is because it's not a monotonic function, which we gather from the table. On the interval [-2, 3], f(x) decreases from 7 to -2, but on the following interval [3, 4], f(x) increases to 3 and becomes positive once more. At best, f(x) is locally invertible.
1. b. This is impossible to determine... We can't know g(4) without knowing for which x we have f(x) = 4, but that's not included in the table.
1. c. The slope of the tangent line to f(x) at x = -2 is f'(-2) = -4.5. Then the equation of the tangent to f(x) at x = -2 - which means it passes through the point (-2, f(-2)) = (-2, 7) - is
y - 7 = -4.5 (x - (-2)) ⇒ y = -4.5x + 16
2. Given f(x) = arctan(3x) - ln(1 + 9x²), compute the derivative:
f'(x) = 3/(1 + (3x)²) - 18x/(1 + 9x²)
f'(x) = (3 - 18x)/(1 + 9x²)
Find the critical points of f(x), where f'(x) = 0 or doesn't exist. Only the first case can happen since the denominator is never zero. So
3 - 18x = 0 ⇒ x = 1/6
is the only critical point.
Also compute the second derivative, then check its sign at the critical point:
f''(x) = (-18 - 54x + 162x²) / (1 + 9x²)²
f''(1/6) = -72/5 < 0
The negative sign indicates that x = 1/6 is the site of a local maximum of
f(1/6) = arctan(1/2) - ln(5/4) ≈ 0.241
Also check the values of f(x) at the endpoints of the interval [0, 1]:
f(0) = 0
f(1) = arctan(3) - ln(10) ≈ -1.054
So, the maximum value of f(x) on [0, 1] is f(1/6) ≈ 0.241.
