Known function f (x) = (e x) (ax+b)-x? -4x, the tangent equation of the image at point (0, f(0)) is y = 4x+4;

( 1)。 Find the values of a and b; (2)。 The monotonicity of f(x) is discussed and the maximum value of f(x) is found.

Solution: (1). F' (x) = (e x) (ax+b)+AE x-2x-4, f '(0)=b+a-4=4, so a+b=8..........( 1).

The tangent equation of(0, f (0)) is y = f' (0) (x-0)+f (0) = 4x+f (0) = 4x+4; So f (0) = b = 4;

Substitute the value of b into equation (1) to get a = 4；; So a=4 and b=4.

(2)。 f(x)=(e^x)(4x+4)-x？ -4 times

Let f' (x) = (EX) (4x+4)+4EX-2x-4 = (EX) (4x+8)-2x-4 = 4 [(EX) (X+2)]-2 (X+2) = 2.

So I have to stay in x? =-2; x？ = ln( 1/2)=-LN2； When -∞ < X≦-2, and [-ln2,+∞) is f '(x)≧0, that is, f(x) is in these two intervals.

Monotonic increase; When -2≦x≦-ln2, f '(x)≦0, that is, f(x) monotonically decreases in this interval. So x? Is it the maximum point, x? This is a tiny point.

The maximum value of f(x) =f(-2)=-4 (1/e? )+4;

Minimum value = f (-ln2) = [e (-ln2)] (-4ln2+4)-ln? 2+4ln2=( 1/2)(4-4ln2)-ln？ 2+4ln2=2( 1-ln2)-ln？ 2+4ln2

=2-(ln？ 2)+2ln2。