AI SL · 12.9
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From the graph of f to the graph of f′
12.9.2
Drag the orange point along the graph of f. Each place you stop leaves a dot on the lower graph at the height of the gradient. Trace f′ yourself, then press Show f′ to check.
Drag the point along f.
The graph shows the straight line y = f(x) through (0, −3) and (2, 1). Write down f′(x).
Have a go first. Then open the working one line at a time.
\(\dfrac{1-(-3)}{2-0}=2\)
A straight line has the same gradient everywhere.
\(f'(x)=2\)
The graph of f′ is a horizontal line at height 2.
y-intercept 2; no x-intercept
f′ is never zero, so f has no turning point.
The graph shows y = f(x), a parabola with its vertex at (2, −1). The graph of f′ is a straight line. Where does it cross the x-axis?
Have a go first. Then open the working one line at a time.
Gradient is zero at \(x=2\)
So the graph of f′ meets the x-axis at x = 2.
\(x<2\): f falls, \(f'<0\). \(x>2\): f rises, \(f'>0\)
f′ is below the axis, then above.
A straight line with positive gradient through (2, 0)
It is \(f'(x)=2x-4\); it meets the y-axis at −4.
The graph shows y = f(x), a cubic with a local maximum at (−1, 5) and a local minimum at (3, −27). Describe the graph of f′.
Have a go first. Then open the working one line at a time.
\(f'(x)=0\) at \(x=-1\) and \(x=3\)
Turning points of f are x-intercepts of f′.
\(x<-1\): f rises, so \(f'>0\). \(-1<x<3\): f falls, so \(f'<0\). \(x>3\): \(f'>0\)
Above the axis, below, above.
An upward parabola through (−1, 0) and (3, 0)
A cubic gives a quadratic gradient. Above, below, above is a U shape.
The graph shows y = f(x), a quartic with minima at (−1, 0) and (1, 0) and a maximum at (0, 1). How many times does the graph of f′ cross the x-axis?
Have a go first. Then open the working one line at a time.
\(x=-1,\ 0,\ 1\)
Three turning points, so f′ has three x-intercepts.
\(f'<0\) for \(x<-1\); \(f'>0\) for \(-1<x<0\); \(f'<0\) for \(0<x<1\); \(f'>0\) for \(x>1\)
Read the direction of f in each interval.
A cubic through \((-1,0)\), \((0,0)\), \((1,0)\) that starts low and ends high
A quartic gives a cubic gradient.
The graph shows y = f(x) for 0 ≤ x ≤ 360, with a maximum at (90, 3) and a minimum at (270, −3). f′ meets the x-axis at x = 90 and where else?
Have a go first. Then open the working one line at a time.
\(x=90\) and \(x=270\)
f′ is zero at both.
f rises on \(0\) to \(90\), falls \(90\) to \(270\), rises \(270\) to \(360\)
So f′ is positive, negative, positive.
A wave that starts at a maximum, crosses at 90, has a minimum at 180, crosses at 270
f is steepest at 0, 180 and 360 (where it crosses its middle), so f′ has its largest and smallest values there. Only the shape matters here.
The table shows f(x) = xᵃ and its gradient function f′(x) for a = 1, 2, 3. Predict f′(x) for a = 4: it is □x³. Find □.
Have a go first. Then open the working one line at a time.
x² → 2x, x³ → 3x²
The old power comes to the front and the power drops by one.
x⁴ → 4x³
The graph of f′ for a = 4 is a cubic through the origin, as in the quartic example.
\(f(x)=x^a\Rightarrow f'(x)=ax^{a-1}\)
You will use this rule soon. For now, notice it.