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).