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Senior one is required 1 math problem.
1, the minimum value of quadratic function Y=f(x) is 4, and f(0)=f(2)=6. Find the analytical formula of f(x).

Solution: f(0)=f(2)=6, which means that the symmetry axis is x =1; And the minimum value is 4, so the vertex is (1, 4). Set the vertex.

y=a(x- 1)? +4; Substituting (0, 6) points gives a=2.

2. Given that f(2x+ 1)=3x+2, find f(5).

Solution: Because f(2x+ 1)=3x+2.

So f (5) = f (2× 2+1) = 3× 2+1= 7.

3. It is known that f(x) is a linear function, which satisfies 3f(x+1)-2f (x-1) = 2x+17, so we can get the analytical formula of f (x).

Solution: let f(x)=ax+b then f (x+1) = a (x+1)+b f (x+1) = a (x-1)+b.

Substitute the original formula 3a (x+1)+3b-2a (x-1)-2b = 2x+17.

ax+5a+b=2x+ 17

Comparing the coefficients of the left and right sides, so a=2 5a+b= 17, we get a=2 and b = 7.

4. Given the functions f(x)= x+2 and g(x)=5x+2, find f(3), f(a+ 1) and f(g(x)).

And find the domain of the function y=f(g(x)).

Solution: f(3)=√(3+2)=√5.

f(a+ 1)= √( a+ 1+2)= √( a+3)

f(g(x))= f(5x+2)= √( 5x+2+2)= √( 5x+4)

F(g(x))=√(5x+4), so 5x+4≥0, so the domain is [-4/5, +∞).