Given f (x) = √ x (x-a)

The derivative of f(x) f' (x) = (2x-a)/2 √ x (x-a),

Let f' (x) = (2x-a)/2 √ x (x-a) = 0,

It is known that x=a/2 and x≠a and x≠0.

When a>0, the domain of f(x) is x≥a∪x≤0.

X∈(-∞, 0] monotonically decreases.

X∈[a, +∞) increases monotonically.

When a < 0, the domain of f(x) is x ≤ a and x ≥ 0.

X∈(-∞, a] monotonically decreases.

X∈[0, +∞) increases monotonically.

When a=0, f (x) = 0;

A and g(a) are the minimum values of f(x) in the interval [0,2].

It is known that a≥0, from the monotone interval mentioned above, f(x) monotonically increases at x∈[a, +∞].

That is to say, (x) monotonically increases at x ∈ [0 0,2].

It is known that g(a)=f(0)=0.

2. Derive f(x) to get lnx+ 1=0.

Let the derivative be zero and x = e (- 1)

X greater than E (- 1) is a increasing function, and less than E (- 1) is a decreasing function.

T is discussed below.

When t is greater than e (- 1), f(t+2) is the largest.

When t+2 is less than e (- 1), f(t) is the largest.

When E (- 1) is between t and t+2, compare f(t) and f(t+2).