A tangent line to a curve is the best linear approximation to the curve near the point of tangency.

Suppose that the line y = 3(x-1) + 2 is tangent to the graph of y = f(x) at the point (1, f(1)),
and that the line y = -2(x-1) + 3 is tangent to the graph of y = g(x) at the point (1, g(1)).
 

• What's f(1)?
• What's g(1)?
• What's f'(1)?
• What's g'(1)?

If we write L(x) = 3(x-1) + 2 and M(x) = -2(x-1) + 3, then L should be a good approximation to f near x=1 and M should be a good approximation to g near x=1.

• Find a good approximation to the function h near x=1, where h(x) = 5f(x).
• What's h(1)?
• What's h'(1)?

• Find a good approximation to the function F near x=1, where F(x) = f(x) + g(x).
• What's F(1)?
• What's F'(1)?

 
• Find a good approximation to the function G near x=1, where G(x) = f(x) - g(x).
• What's G(1)?
• What's G'(1)?

 
• Find a good approximation to the function H near x=1, where H(x) = f(x) * g(x).
• The most obvious answer to the preceding question isn't a linear function. Find a linear approximation to your approximation. One way to do this is to assume your approximation has a formula of the form m*(x-1)+b. Set this equal to your original approximation. Then (1) set x = 1 and solve for b, (2) substitute this value for b, (3) solve the resulting equation for m, and (4) set x = 1 to find a constant value for m.
• What's H(1)?
• What's H'(1)?

 
• Find a good approximation to the function K near x=1, where K(x) = f(x)/g(x).
• The most obvious answer to the preceding question isn't a linear function. Find a linear approximation to your approximation.
• What's K(1)?
• What's K'(1)?