Since f(1) = 1, the point we want is (1,1).Since we found earlier that f'(1) = 2, 2 is the slope we want.
We want a line of the form
y = mx + b whose slope is 2, so m = 2: y = 2x + b.
We also want (1,1) to be on the line, so
1 = 2(1) + b.
Solving, we find b = -1. This means the equation for the tangent line to f at 1 is
y = 2x-1.
To check this answer, we graph the function f(x) = x2and the line y = 2x - 1 on the same graph:
Since the line bounces off the curve at x = 1, this looks like a reasonable answer.
When finding equations for tangent lines, check the answers. Stick both the original function and the tangent line in the calculator, and make sure the picture looks right. If we find something like this, we know we've made a mistake somewhere:
If we find something that looks like a tangent line and quacks like a tangent line, there's a good chance we've correctly found the tangent line.
Since we haven't discussed the shortcuts for finding derivatives yet, these exercises will require derivatives we've already found.
We can find the derivatives again for practice or we can go look them up.