If we remember two things, we can write the equation for the tangent line to f at a given the formula for f and the value a where we want the tangent line to go.
- We can calculate f(a).
- We can calculate f ' (a).
We can write the equation of a line given a point and a slope, so once we have a point and a slope for the tangent line we're all set to go.
We've left out one thing in this discussion of tangent lines: the magic formula. Most textbooks have a magic formula that produces the equation for the tangent line.
We have lots of reasons for leaving this formula out. We don't need it. It takes extra memory that could be better spent remembering the limit definition of the derivative. It can be confusing.
However, for the sake of completeness, we'll show the magic formula. But we're doing it our way.
There's one special case that neither finding the equation of a line nor knowing the magic formula will help with. If f ' (a) is undefined and infinite, then we have a vertical tangent line.
The equation of such a tangent line is x = a like any vertical line. Knowing this will probably be more important for finding derivatives of parametric functions than it will be right now.
We can also work backwards to figure out information about the function given information about its tangent line.
Practice:
Find the tangent line to f(x) = x2 at x = 1. | |
We need a point and a slope. 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) = x2 and 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. | |
Find the equation for the tangent line to f at a. | |
There are no numbers, but don't panic. This will work the same way as all the other problems. We need a point and a slope. Since we don't have a formula, we can't figure out what f(a) is, but we know the point (a, f(a)) is the one we want. Since we don't have a formula, we can't figure out what f'(a) is, but we know f'(a) is the slope we want. We want a line of the form y = mx + b whose slope is f'(a), so m = f'(a): y = f'(a)x + b. We also want (a, f(a)) to be on the line, so f(a) = f'(a)(a) + b. Solving, we find b = f(a)-f'(a)a. This means the equation for the tangent line to f at a is
y = f'(a)x + (f(a)-f'(a)a). This is a little messy. We can make it prettier by rearranging. First, remove the unnecessary parentheses: y = f'(a)x + f(a)-f'(a)a Then swap the second and third terms: y = f'(a)x-f'(a)a + f(a) Pull out f'(a) from the first two terms: 
And there we are. The magic formula says the equation for the tangent line is y = f'(a)(x - a) + f(a). We might also see it written y = f(a) + f'(a)(x - a), which is the same thing with the terms switched. Since we've given in and explained the magic formula, we should probably show how to use it, too. Here, we aren't as nice. We'll need to find the derivatives from scratch, using the sneaky methods. | |
Let f(x) = 3 + x + x2. Use the magic formula to find the tangent line to f at a = 1.
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We know a = 1, so f(a) = f(1) = 3 + 1 + 1 = 5. Now we need to find f'(1). We have a = 1. 
Since f(1) = 5, we have 
Now we have all the pieces: a = 1, f(a) = 5, and f'(a) = 3. We put them all into the formula y = f(a) + f'(a)(x-a) to find y = 5 + 3(x - 1). We may or may not want to "simplify" this to find y = 3x + 2. As usual, we can check the answer by graphing the original function and the tangent line. In the previous exercise, we prefer the answer y = 5 + 3(x - 1) because we can look at that function and tell by the way it looks that it came from the magic formula. It may be necessary to simplify. y = 3x + 2, which is admittedly easier to graph. | |
The picture below shows a function and its tangent line at 2: 
What is f'(2)?
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f'(2) is the slope of the tangent line to f at 2. Since we have two points on the tangent line, we can find its slope: 
This means f'(2) = 0.5. | |
For the given function f andvalue a, find the tangent line to f at a.
Answer
- f(x) = x2 + 1, a = 1 We need a point and a slope. Since a = 1 and f(1) = 2, the point we want is (1,2). We found earlier that f'(1) = 2. We want a line
y = mx + b
whose slope is f'(1) = 2, so
y = 2x + b.
We want the point (1,2) to be on the line, so
2 = 2(1) + b.
Solving, we find b = 0 so the equation of the tangent line is
y = 2x.
We can check this with a picture:
For the given function f and value a, find the tangent line to f at a.
Answer
- f(x) = x2 + 1, a = 0We need a point and a slope. Since a = 0 and f(0) = 1, the point we want is (0,1). We found earlier that f'(0) = 0. We want a line
y = mx + b
whose slope is f'(0) = 0, so
y = (0)x + b = b.
We want the point (0,1) to be on the line, so
1 = b.
This means the equation of the tangent line is
y = 1.
We can check this with a picture:
For the given function f and value a, find the tangent line to f at a.
Answer
We need a point and a slope. Since a = 1 and f(1) = 1, the point we want is (1,1). We found earlier that f'(1) = 3. We want a line
y = mx + b
whose slope is f'(1) = 3, so
y = 3x + b.
We want the point (1,1) to be on the line, so
1 = 3(1) + b.
Solving, we find b = -2 so the equation of the tangent line is
y = 3x-2.
We can check this with a picture:
For the given function f and value a, find the tangent line to f at a.
Answer
We need a point and a slope. Since a = -1 and f(-1) = 0, the point we want is (-1,0). We found earlier that f'(-1) = 2. We want a line
y = mx + b
whose slope is f'(-1) = 2, so
y = 2x + b.
We want the point (-1,0) to be on the line, so
0 = 2(-1) + b.
Solving, we find b = 2 so the equation of the tangent line is
y = 2x + 2.
We can check this with a picture:
For the given function f and value a, find the tangent line to f at a.
Answer
We need a point and a slope. Since a = 1 and 
, the point we want is 
. We found earlier that
. We want a line
y = mx + b
whose slope is 
, so
We want the point 
to be on the line, so
Solving, we find 
so the equation of the tangent line is
We can check this with a picture:
For the function f and value of a, use the magic formula to find the tangent line to f at a. We will need to calculate derivatives from scratch.
Answer
Since a = 4, f(4) = 2(4) + 3(4)2 = 56. Now we need to find f'(4).
We put this all into the formula
y = f(a) + f'(a)(x - a)
to find
y = 56 + 26(x - 4).
This can be simplified to
y = 26x - 48.
For the function f and value of a, use the magic formula to find the tangent line to f at a. We will need to calculate derivatives from scratch.
Answer
Since a = -2, f(a) = f(-2) = 2(-2)3 = -16.
To find f'(-2), we have
We leave it for you to check that
f(-2 + h) = 2h3 - 12h2 + 24h - 16.
Then
We put this all into the formula
y = f(a) + f'(a)(x - a)
to find
y = -16 + 24(x- (-2)) = -16 + 24(x + 2).
This can be simplified to
y = 24x + 32.
For the function f and value of a, use the magic formula to find the tangent line to f at a. We will need to calculate derivatives from scratch.
Answer
Since a = 1, f(a)= f(1) = 1.
We need to find f'(1):
We put this all into the formula
y =f(a)+f'(a)(x-a)
to findy = 1 + (-2)(x - 1)which can be simplified toy = -2x + 3.
The graph shows a function f and a line that is tangent to f at a. For the graph determine a, f(a), and f'(a) (a refers to the x-value at which the line is tangent to f).
Answer
- The line is tangent to f at 1, so a = 1. Since the point (1,0.5) is on the graph of f, f(a) = 0.5. To find f'(a) we find the slope of the tangent line to f at a.
