- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
The definition of the derivative practice question 13
Use the definition of the derivative to find the slope of the tangent line to at . Then find the -intercept of that tangent line, and enter it.
Rules such as the power rule earn no marks on this question.
First the slope, from the definition, then the line. In the difference quotient for , what must you do before can cancel?
Set up and combine the two fractions over the common denominator .
The tangent line passes through with the slope you found. Write it in point-slope form and set .
A number.
Theory UBC Math 100 §3.3CLP-1 §2.2Key results
Worked solution
Strategy
Get the slope from the difference quotient, combining the fractions until the h cancels. The point of tangency is (2, 1/2); point and slope give the line, and its value at x = 0 is the intercept.
Working
Common denominator 2(2 + h). The numerator is a multiple of h, as it must be.
Cancel h, then let h approach 0.
Point-slope form through (2, 1/2), simplified.
Set x = 0.
Answer
The answer is the tangent line is y = −x/4 + 1, with y-intercept 1.
Check
The line y = 1 − x/4 meets y = 1/x only at x = 2, since x(1 − x/4) = 1 rearranges to (x − 2)² = 0, a double root.
Where this goes wrong
Giving 1/2, the height of the curve at x = 2, as the intercept. The point (2, 1/2) is on the tangent line; the intercept is where the line meets x = 0.
Leans on
Point-slope form of a line, from Functions, graphs and inverses
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