Tangent line problems
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- Dustin Riley
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1 You will find lots of practice problems and homework problems that simply ask you to differentiate. The following examples are to illustrate some of the types of tangent line problems that you may come across.
2 Find an equation of the tangent line to f (x) = x + 3 at x = 4.
3 Find an equation of the tangent line to f (x) = x + 3 at x = 4. To write the equation of the tangent line we will need a point and a slope.
4 Find an equation of the tangent line to f (x) = x + 3 at x = 4. To write the equation of the tangent line we will need a point and a slope. point: At x = 1, f (4) = = 5, and so the point is (4, 5)
5 Find an equation of the tangent line to f (x) = x + 3 at x = 4. To write the equation of the tangent line we will need a point and a slope. point: At x = 1, f (4) = = 5, and so the point is (4, 5) slope: We differentiate to get f (x) = 1 x 1/, and then m = f (4) = 1 ( 1 )( 1 ) ( 1 )( 1 (4) 1/ = 4 = = ) 1 4
6 Find an equation of the tangent line to f (x) = x + 3 at x = 4. To write the equation of the tangent line we will need a point and a slope. point: At x = 1, f (4) = = 5, and so the point is (4, 5) slope: We differentiate to get f (x) = 1 x 1/, and then m = f (4) = 1 ( 1 )( 1 ) ( 1 )( 1 (4) 1/ = 4 = = ) 1 4 Therefore the equation of the tangent line is y 5 = 1 (x 4). 4
7 Find an equation of the tangent line to f (x) = 3x that is perpendicular to y = x + 5.
8 Find an equation of the tangent line to f (x) = 3x that is perpendicular to y = x + 5. Again, we need a point and a slope. Since we want a tangent line that is perpendicular to y = x + 5, which has a slope of. Since perpendicular lines have negative reciprocal slopes, we want a slope of 1. So we have our slope, in this problem we have to find the point.
9 Find an equation of the tangent line to f (x) = 3x that is perpendicular to y = x + 5. Again, we need a point and a slope. Since we want a tangent line that is perpendicular to y = x + 5, which has a slope of. Since perpendicular lines have negative reciprocal slopes, we want a slope of 1. So we have our slope, in this problem we have to find the point. To find the point, we need to determine where on the graph of f (x) = 3x is there a tangent line with slope 1. How do we do this?
10 Find an equation of the tangent line to f (x) = 3x that is perpendicular to y = x + 5. Again, we need a point and a slope. Since we want a tangent line that is perpendicular to y = x + 5, which has a slope of. Since perpendicular lines have negative reciprocal slopes, we want a slope of 1. So we have our slope, in this problem we have to find the point. To find the point, we need to determine where on the graph of f (x) = 3x is there a tangent line with slope 1. How do we do this? Solve f (x) = 1!
11 Find an equation of the tangent line to f (x) = 3x that is perpendicular to y = x + 5. f (x) = 6x, so we need to solve 6x = 1 1 which gives us x = 1. This is the x-value of the point on the graph of f (x) = 3x that has a tangent line with slope 1.
12 Find an equation of the tangent line to f (x) = 3x that is perpendicular to y = x + 5. f (x) = 6x, so we need to solve 6x = 1 1 which gives us x = 1. This is the x-value of the point on the graph of f (x) = 3x that has a tangent line with slope 1. ( ) ( ) To get the y-value of the point, f 1 1 = = Then the equation of the line is, y + 95 ( 48 = 1 x + 1 ). 1
13 Determine the points (if any) on the graph of f (x) = x 3/ x where f has a horizontal tangent line.
14 Determine the points (if any) on the graph of f (x) = x 3/ x where f has a horizontal tangent line. First, think about what type of slope a horizontal line has. If you can t remember draw a horizontal line, then choose two points on the line and use them to calculate the slope.
15 Determine the points (if any) on the graph of f (x) = x 3/ x where f has a horizontal tangent line. First, think about what type of slope a horizontal line has. If you can t remember draw a horizontal line, then choose two points on the line and use them to calculate the slope. Horizontal lines have slope 0. This problem is asking if f (x) = 0 has any solutions. Calculate f (x) then click next.
16 Determine the points (if any) on the graph of f (x) = x 3/ x where f has a horizontal tangent line. First, think about what type of slope a horizontal line has. If you can t remember draw a horizontal line, then choose two points on the line and use them to calculate the slope. Horizontal lines have slope 0. This problem is asking if f (x) = 0 has any solutions. Calculate f (x) then click next. f (x) = 3 x 1/, now we solve f (x) = 0.
17 Determine the points (if any) on the graph of f (x) = x 3/ x where f has a horizontal tangent line. 3 x 1/ = 0 3 x = 4 x = 3 x = 16 9 There ( is one ) point on f (x) that has a horizontal tangent line, 16 9, 3 7. (To get y-value, plug x = 16 9 into f (x) = x 3/ x.)
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