Let f(x)=-4X^2+8X+13 Find g'(3) Show that the linear approximation to g(3+△x) always gives an answer which is too large,regardless of whether △x is positive or negative.Interpret your answer geometrically by drawing graph of g and its tangent lin

1.given the function f(x) = x*2 -4x -5,find the values of:a) f(k-2) b) f(x*2) - x*2 f(1)2.given two functions f(x) = x*2 -3 and g(x) = 3x-4,find the values of:a) f(k+1) -g(k-1) b) f(g(x)) c) g(f(x))3.let f(x) = 1/1-x,finda) f(f(x)) b) f(f(f(x))) c) f 1.given the function f(x) = 2x*2 +5x -7,find the values of:a) f(-2) b) f(4) c) f(x*2) d) (f(x))*2 e) f(x*2) -xf(x)2.let f(x) =1/2 (2*x (符号是二的x次方) +2*-x (二的-x次方))show that for any real values of x and y,f(x+y) +f (x-y) = 2f(x)f(y matlab解决约束非线性规划问题myfun.mfunction f=myfun(x)f=x(1)*x(13)+x(2)*x(14)+x(3)*x(15)+x(25)+1.697*(x(4)*x(16)+...x(5)*x(17)+x(6)*x(18)+x(26))+0.575*(x(7)*x(19)+x(8)*x(20)...+x(9)*x(21)+x(27))+0.723*(x(10)*x(22)+x(11)*x(23)+x(12)*x(24)); matlab解决约束非线性规划问题myfun.mfunction f=myfun(x)f=x(1)*x(13)+x(2)*x(14)+x(3)*x(15)+x(25)+1.697*(x(4)*x(16)+...x(5)*x(17)+x(6)*x(18)+x(26))+0.575*(x(7)*x(19)+x(8)*x(20)...+x(9)*x(21)+x(27))+0.723*(x(10)*x(22)+x(11)*x(23)+x(12)*x(24)); f(x)值域 [ 3/8，4/9] y=f(x)+√1-2f(x) f(1+1/x)=x/1-x^2 求f(x) f(x)=x^2+x (x 已知f(x)=x^4-x^3-7x^2+13x-6.x-1、x-2、x+3都是f(x)的一个因式,求证f(x)能被(x-1)(x-2)(x+3)整除. F(X)=2X^3+6X^2-8X计算：（1）F(X)•(-2X);(2)2F(X)-4X f(X)=f(X+2)(x 设f（x)满足f(-x)=-f(x),f(x)=f(4-x),x∈[0,2)时,f(x)=x,则f(11.5)等于? f(x)=5x^2-4x-3f(x)=-3x^2+5x-8配方 已知f(x)=-16x^4+8x^3-12x^2,求f(x)/[(-1/4)x] f(x)=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.1/(x+9)(x+10) 求f(8) 若f(x)=2f(2-x)-x^2+8x-8,求f(x) f(x+1)+f(x-1)=2x²-4x,求f（x）? f(x+1)+f(x-1)=4x^3-2x求f(x) f(x)=-2x+8,那么f(3x^+4x+7)=多少 f(x)=min{x^2-2,x,4-x}