Verified answer

f(x)=6x-x^(2)

The difference quotient is the increment of the value of a function divided by the increment of the independent variable.

(f(x+h)-f(x))/(h)

Replace the variable x with x+h in the expression.

f(x+h)=6(x+h)-(x+h)^(2)

Multiply 6 by each term inside the parentheses.

f(x+h)=(6x+6h)-(x+h)^(2)

Squaring an expression is the same as multiplying the expression by itself 2 times.

f(x+h)=(6x+6h)-((x+h)(x+h))

Multiply -1 by each term inside the parentheses.

f(x+h)=(6x+6h)-x^(2)-2xh-h^(2)

Reorder the polynomial 6x+6h-x^(2)-2xh-h^(2) alphabetically from left to right, starting with the highest order term.

f(x+h)=-x^(2)+6x-2xh-h^(2)+6h

Substitute the known value into the formula for the difference quotient.

(f(x+h)-f(x))/(h)=(-x^(2)+6x-2xh-h^(2)+6h-(6x-x^(2)))/(h)

Reorder the polynomial 6x-x^(2) alphabetically from left to right, starting with the highest order term.

(f(x+h)-f(x))/(h)=(-x^(2)+6x-2xh-h^(2)+6h-(-x^(2)+6x))/(h)

Multiply -1 by each term inside the parentheses.

(f(x+h)-f(x))/(h)=(-x^(2)+6x-2xh-h^(2)+6h+(x^(2)-6x))/(h)

Since -x^(2) and x^(2) are like terms, subtract x^(2) from -x^(2) to get 0.

(f(x+h)-f(x))/(h)=(0+6x-2xh-h^(2)+6h-6x)/(h)

Since 6x and -6x are like terms, add -6x to 6x to get 0.

(f(x+h)-f(x))/(h)=(0-2xh-h^(2)+6h)/(h)

Combine all similar terms in the polynomial -x^(2)+6x-2xh-h^(2)+6h+x^(2)-6x.

(f(x+h)-f(x))/(h)=(-2xh-h^(2)+6h)/(h)

Divide each term in the numerator by the denominator.

(f(x+h)-f(x))/(h)=-(2xh)/(h)-(h^(2))/(h)+(6h)/(h)

Remove the common factors that were cancelled out.

(f(x+h)-f(x))/(h)=-2x-(h^(2))/(h)+(6h)/(h)

Reduce the expression -(h^(2))/(h) by removing a factor of h from the numerator and denominator.

(f(x+h)-f(x))/(h)=-2x-h+(6h)/(h)

Remove the common factors that were cancelled out.

(f(x+h)-f(x))/(h)=-2x-h+6

the answer is 6-2x if you take lim(f(x+h)-f(x))/h , as h---> 0