It is known that the functions f x ax 4 3ax b, f 1 2, f 1 1 1 find the analytic formula of f x

1)∵f(x)=ax²－4/3ax＋b,f(1)=2,f'(1)=1f(1)＝a－4a/3＋b＝﹣a/3＋b＝2f'(x)＝2ax－4a/3 ∴f'(1)＝2a－4a/3＝2a/3＝1 ∴a＝3/2 b＝5/2

f(x)＝3/2 x²－2x＋5/2

2)∵f'(1)=1 f(x) The tangent slope at (1,2) is 1, and the tangent equation is: y x m

Over (1,2) 1 m 2 m 1

The tangent equation for f(x) at (1,2) is: y x 1

1) According to the question, f(1)=a*1-4 3a*1+bb-a 3=2

f'(x)=2ax-4/3a

So f'(1)=2a-4 3a=2 3a=1 is solved to obtain a=3 2, b=5 2

f(x)=3/2x^2-2x+5/2

2), f(x)=3 2x 2-2x+5 2 so f'(x)=3x-2

Slope k=f'(1)=1

The tangent equation yields y=kx+b=x+b

Substituting (1,2).

We get b=1, so the tangent equation is y=x+1

and x-y+1=0

Summary. a 2 is known to f(-1)=4 and brings x -1 into the equation to get a+2 4 so a 2

The function f(x)=ax -2x is known, and if f(-1)=4, then a= is to be parsed. Okay.

Process. Hmmm.

Parse: I'll send it to you now.

What about the process. Okay thank you.

Good. a 2 is known to f(-1)=4 and brings x -1 into the equation to get a+2 4 so a 2

A 2 is not a 3ok

Answer: f(2)=-18

First, find the unknown a in f(x)=ax-4. Method: In the question, Qingnai has given f(-1)=3, which means that when x=-1, f(x)=3, -1 and 3 are substituted into f(x)=ax-4, and a=-7 is obtained.

Substituting the modulo index a=-7 into f(x)=ax-4 gives f(x)=-7x-4

Next, find f(2). Method: Substituting x=2 into the formula we finally derived: f(x)=-7x-4, we get f(2)=-7 2-4=-14-4=-18

Knowing that the function $f(x)=ax-4$ and $f(-1)=3$, we can use the known condition to argue the value of silver out $a$: finger stove.

f(-1)=a times (-1)-4=3$$ solution is bent to $a=1$.

So, $f(x)=x-4$, then $f(2)=2-4=-2$.

So, $f(2) = -2$.

Depending on the conditions.

a＋b＝0，①

4a－2b＝3，②

The solution is a 1 2, b 1 2, so the analytic formula of the function is y faction (x x) to change the calendar 2. Dust years.