AI SL · 12.9
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From the graph of f′ back to f
12.9.3
The graph shows y = f′(x), a straight line through (0, 3) and (3, 0). Write the values of x for which f is increasing.
Have a go first. Then open the working one line at a time.
f increasing \(\Leftrightarrow f'(x)>0\)
Look for where the graph of f′ is above the x-axis.
\(f'(x)=0\) at \(x=3\)
This is a stationary point of f.
The line is above the axis for \(x<3\)
It slopes down, so it is positive on the left.
f is increasing for \(x<3\)
The graph of f′ is not the graph of f. Only its sign matters here.
The graph shows y = f′(x), a parabola meeting the x-axis at x = 1 and x = 5 and with its lowest point at (3, −4). Write the values of x for which f is increasing.
Have a go first. Then open the working one line at a time.
\(f'(x)=0\) at \(x=1\) and \(x=5\)
f has stationary points here.
\(f'(x)>0\) for \(x<1\) and for \(x>5\)
The parabola opens upwards, so it is above the axis outside the zeros.
Increasing: \(x<1\) or \(x>5\)
Between 1 and 5, f′ is negative, so f is decreasing.
The graph shows y = f′(x). It has a vertical asymptote x = 1 and a horizontal asymptote y = 1 (dashed), and meets the x-axis at x = −1. Write the values of x for which f is increasing.
Have a go first. Then open the working one line at a time.
For \(x<-1\), \(f'(x)>0\)
The left branch starts above the line y = 1, then falls and crosses the x-axis at x = −1.
For \(-1<x<1\), \(f'(x)<0\)
The graph is below the axis and drops towards the asymptote.
For \(x>1\), \(f'(x)>0\)
The right branch is above the axis everywhere. At x = 1 itself f′ is not defined.
Increasing: \(x<-1\) or \(x>1\)
The asymptote is where the right-hand interval starts.
The graph shows y = f′(x), a cubic that meets the x-axis at x = −2, 0 and 3. Write the values of x for which f is increasing.
Have a go first. Then open the working one line at a time.
\(x=-2,\ 0,\ 3\)
f has three stationary points.
\(x<-2\): below. \(-2<x<0\): above. \(0<x<3\): below. \(x>3\): above
Read the graph interval by interval.
Increasing: \(-2<x<0\) or \(x>3\)
Only the intervals where f′ is above the axis.
The graph shows y = f′(x), the line through (0, −8) and (4, 0). Find the x-coordinate of the stationary point of f, and say whether it is a maximum or a minimum.
Have a go first. Then open the working one line at a time.
\(f'(x)=0\) at \(x=4\)
The stationary point is at x = 4.
\(x<4\): \(f'<0\), f falls. \(x>4\): \(f'>0\), f rises
Falling then rising.
A U shape with its lowest point at x = 4: a minimum
Mark the stationary point on the sketch. A constant f′, such as f′(x) = 3, would give a straight line of gradient 3 with no stationary point.
The graph shows y = f′(x) for x > 0. It is a decreasing curve that gets closer to the x-axis. It passes through (1, 4), (2, 2) and (4, 1). Find the gradient of f at x = 2, and describe how the gradient of f changes as x increases.
Have a go first. Then open the working one line at a time.
\(f'(2)=2\)
The height of the f′ graph is the gradient of f.
\(f'(x)>0\) for all \(x>0\)
So f is increasing for all x > 0, with no stationary point.
f′ decreases towards 0
The gradient of f is positive but gets smaller: f keeps rising but flattens out.