Estimating the Slope of a Curve Using Tangent Lines
The slope of a curve at a particular point represents the instantaneous rate of change of the function at that point. Since curves are not straight lines, their slope varies along the curve.
To estimate the slope at a point, draw a tangent line to the curve at that point. A tangent line just touches the curve without cutting through it nearby.
Once the tangent is drawn, calculate its gradient by choosing two points on the tangent line and using the formula: gradient = (change in y) / (change in x). This gives an approximate slope of the curve at that point.
- Draw a tangent line carefully at the point of interest.
- Select two points on the tangent line, preferably far apart for accuracy.
- Calculate gradient = (y₂ - y₁) / (x₂ - x₁).
- This gradient estimates the slope of the curve at the point.
Differentiating Functions of the Form axⁿ
Differentiation provides an exact method to find the slope of a curve defined by a function at any point.
For functions of the form f(x) = axⁿ, where 'a' is a rational constant and 'n' is a non-negative integer, the power rule applies:
The derivative f'(x) = a × n × xⁿ⁻¹, where xⁿ⁻¹ means x raised to the power (n - 1).
- Example: Differentiate f(x) = 3x⁴.
- Apply the power rule: f'(x) = 3 × 4 × x³ = 12x³.
- This derivative function gives the slope of the curve at any x-value.
Differentiating Sums of up to Three Functions
When a function is a sum of terms each in the form axⁿ, differentiate each term separately and then add the results.
This is because differentiation is a linear operation.
For example, if f(x) = 2x³ + 5x² - 4x, differentiate each term: 2x³ → 6x², 5x² → 10x, -4x → -4.
- Example: Differentiate f(x) = x³ + 2x² + 7.
- Derivative: f'(x) = 3x² + 4x + 0 = 3x² + 4x.
- Constant terms differentiate to zero.
Finding Stationary Points
Stationary points occur where the slope of the curve is zero, meaning the derivative equals zero.
To find stationary points: 1) Differentiate the function to get f'(x). 2) Solve f'(x) = 0 for x.
Each solution gives an x-coordinate of a stationary point. Substitute back into the original function to find the corresponding y-coordinate.
- Example: Find stationary points of f(x) = x³ - 3x² + 4.
- Step 1: Differentiate: f'(x) = 3x² - 6x.
- Step 2: Solve 3x² - 6x = 0 → 3x(x - 2) = 0 → x = 0 or x = 2.
- Step 3: Find y-values: f(0) = 0 - 0 + 4 = 4; f(2) = 8 - 12 + 4 = 0.
- Stationary points at (0, 4) and (2, 0).
Classifying Stationary Points as Maxima or Minima
Once stationary points are found, determine whether each is a maximum, minimum, or neither (such as a point of inflection).
Three methods are commonly used: 1) Sketching the curve accurately around the stationary point to observe its shape. 2) The second derivative test:
Calculate the second derivative f''(x). Substitute the stationary point's x-value into f''(x). If f''(x) > 0, the point is a minimum (curve is concave up). If f''(x) < 0, the point is a maximum (curve is concave down). 3) Gradient sign test: Examine the sign of f'(x) just before and after the stationary point: if gradient changes from positive to negative, it's a maximum; if from negative to positive, it's a minimum.
- Example: For f(x) = x³ - 3x² + 4, stationary points at x=0 and x=2.
- Second derivative: f''(x) = 6x - 6.
- At x=0: f''(0) = -6 < 0 → maximum at (0,4).
- At x=2: f''(2) = 6 → minimum at (2,0).
- Sketch or gradient sign test can confirm these results.
At-a-glance summary
Comparison table
| Step | Detail | Example | Exam Use |
|---|---|---|---|
| Estimating slope | Draw tangent and calculate gradient | Tangent at x=1 on y=x² gives gradient ≈ 2 | Use when exact differentiation is not possible or to check answers |
| Differentiating axⁿ | Apply power rule: d/dx(axⁿ) = a·n·xⁿ⁻¹ | d/dx(5x³) = 15x² | Basic skill for finding slopes and stationary points |
| Differentiating sums | Differentiate each term separately and add | d/dx(x³ + 2x) = 3x² + 2 | Needed for polynomial functions with multiple terms |
| Finding stationary points | Solve f'(x) = 0 | For f(x)=x³-3x²+4, stationary points at x=0,2 | Essential for locating turning points |
| Classifying stationary points | Use sketch, second derivative or gradient test | f''(0)<0 max, f''(2)>0 min for f(x)=x³-3x²+4 | Determines nature of turning points for graph interpretation |
Active recall
Quick check
1. How do you estimate the slope of a curve at a point without calculus?
By drawing a tangent line at that point and calculating the gradient of the tangent line.
2. What is the derivative of f(x) = 4x³?
f'(x) = 4 × 3 × x² = 12x².
3. How do you find stationary points of a function?
Differentiate the function to get f'(x), then solve f'(x) = 0 for x.
4. What does a positive second derivative at a stationary point indicate?
It indicates the stationary point is a minimum.
5. If the gradient changes from positive to negative at a stationary point, what type of point is it?
It is a maximum point.