Find the slope of the tangent line at x=3
F(x)=-2x+1
[-2(3+h)+1] - [-2(3)+1]
÷
h
but it is a straight line of form
y = m x + b
where m is the slope = 2
but oh well, we can take the derivative
{[+2(3+h)+1] - [+2(3)+1] }/h
note + 2 not -
[6 + 2 h +1 - 6 -1 ] / h
2 h/h
2
Good work, Damon, but the slope was in fact -2.
To find the slope of the tangent line at x=3, we need to compute the derivative of the function F(x)=-2x+1.
The derivative of a linear function like F(x)=-2x+1 is simply the coefficient in front of x. In this case, the derivative of F(x) is -2.
Therefore, the slope of the tangent line at x=3 is -2.