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Discussion
In general, decribe the behavior of a function f when
f′(a)>0.
f′(x)<0.
f′(x)=0.
Part (a) (i)
We can think of f′(x) as the slope of f or the rate of change of y with respect to x.
If we consider f′(a) as the slope at a, we can relate to lines, where positive slopes indicate that a line goes up and to the right (or down and to the left).
Similarly, if we think of f′(a) as the rate of change of y with respect to x at a, then a positive derivative indicates that y increases as x increases (or decreases as x decreases).
In either case, a positive derivative tells us that the points of f immediately to the left of a are lower than f(a) and that the points of f immediately to the right of a are above f(a).
Hence, when f′(a)>0, we say f is increasing at a because the value of f increases as we move from left to right.
Part (a) (ii)
We can think of f′(x) as the slope of f or the rate of change of y with respect to x.
If we consider f′(x)<0 as the slope at x, then the tangent line falls as it moves to the right (or rises as it moves to the left).
Similarly, as a rate of change, a negative derivative indicates that y decreases as x increases (or increases as x decreases).
In either case, the points of f immediately to the left of x lie above f(x) and those immediately to the right lie below f(x).
Hence when f′(x)<0, we say f is decreasing at x because the value of f falls as we move from left to right.
Part (a) (iii)
We can think of f′(x) as the slope of f or the rate of change of y with respect to x.
If we consider f′(a)=0 as the slope at a, it corresponds to a horizontal line at that point.
Similarly, if we think of f′(a) as the rate of change of y with respect to x at a, then zero derivative indicates that y neither increases nor decreases as x increases or decreases.
In either case, a zero derivative tells us that the points of f immediately to the left of a are at the same height as f(a) and that the points immediately to the right of a are also at the same height
Hence, when f′(a)=0, we say f is stationary at a, meaning it has a horizontal tangent and is neither increasing nor decreasing there.
Stationary points & Increasing/Decreasing Regions
Checkpoint
Given f′(x)=5x−3, state the intervals on which f is increasing, decreasing, and stationary.
Select the correct option
Discussion
We can better classify stationary points based on the derivative of the function at nearby points.
Suppose f′(a)=0, f′(x)<0 to the left of a, and f′(x)>0 to the right of a. Interpret how (a,f(a)) relates to the surrounding points of f.
Now, suppose f′(a)=0, f′(x)>0 to the left of a, and f′(x)<0 to the right of a. Interpret how (a,f(a)) relates to the surrounding points of f.
Check your findings: it is given that f(x)=x2. Find f′(x). State the stationary point of f, and determine whether f is increasing or decreasing to the left and right of that point. Finally, conclude whether the stationary point is a maximum or minimum.
Part (a)
Since f′(x)<0 for x<a, f is decreasing as x approaches a from the left.
Since f′(x)>0 for x>a, f is increasing as x moves away from a to the right.
It is sufficient to say that f is decreasing as it approaches a from the left and increasing as it moves away from a to the right.
However, it is important to notice that since f decreases toward a, f(a) is below the points of f to the left of a, and since f increases away from a, f(a) is below the points of f to the right of a. Hence, we see that f(a) lies below all of the points of f to its immediate left and right. Therefore, we say f(a) is a local minimum.
Part (b)
Since f′(x)>0 for x<a, f is increasing as x approaches a from the left.
Since f′(x)<0 for x>a, f is decreasing as x moves away from a to the right.
It is sufficient to say that f is increasing as it approaches a from the left and decreasing as it moves away from a to the right.
However, it is important to notice that since f increases toward a, f(a) is above the points of f to the left of a, and since f decreases away from a, f(a) is above the points of f to the right of a. Hence, we see that f(a) lies above all of the points of f to its immediate left and right. Therefore, we say f(a) is a local maximum.
Part (c)
Differentiating gives
Setting f′(x)=0 yields
so the stationary point is
For x<0, f′(x)=2x<0, so f is decreasing to the left of 0. For x>0, f′(x)=2x>0, so f is increasing to the right of 0.
Because f goes from decreasing to increasing at x=0, (0,0) is a local minimum.
Maxima & Minima
Stationary points are often local extrema.
If f′(a)=0, f is decreasing to the left of a (f′(x)<0), and f is increasing to the right of a (f′(x)>0), then (a,f(a)) is a local minimum.
If f′(a)=0, f is increasing to the left of a (f′(x)<0), and f is decreasing to the right of a (f′(x)>0), then (a,f(a)) is a local maximum.
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Optimisation
Optimisation problems require you to find a minimum or maximum value by producing a function f(x), taking its derivative, solving f′(x)=0, and confirming which stationary point(s) are minima or maxima.
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