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Justification using first derivative

Let's take a close look at how the behavior of a function is related to the behavior of its derivative. This type of reasoning is called "calculus-based reasoning." Learn how to apply it appropriately.
A derivative f gives us all sorts of interesting information about the original function f. Let's take a look.

How f tells us where f is increasing and decreasing

Recall that a function is increasing when, as the x-values increase, the function values also increase.
Graphically, this means that as we go to the right, the graph moves upwards. Similarly, a decreasing function moves downwards as we go to the right.
Function f is graphed. The x-axis is unnumbered. The graph consists of a curve. The curve starts in quadrant 3, moves upward, or is increasing, concave down to a point in quadrant 2, moves downward, or is decreasing, concave down until a point in quadrant 4 and then moves downward concave up to a point in quadrant 4, moves upward, or is increasing, concave up, and ends in quadrant 1.
Now suppose we don't have the graph of f, but we do have the graph of its derivative, f.
Function f prime is graphed. The x-axis is unnumbered. The graph consists of U-shaped curve. The curve starts in quadrant 2, moves downward concave up through the negative x-axis to a point in quadrant 4, moves upward concave up through a point on the positive x-axis, and ends in quadrant 1.
We can still tell when f increases or decreases, based on the sign of the derivative f:
  • The intervals where the derivative f is positive (i.e. above the x-axis) are the intervals where the function f is increasing.
  • The intervals where f is negative (i.e. below the x-axis) are the intervals where f is decreasing.
The graph of function f prime has 3 portions highlighted. The portion of the graph that moves downward in quadrant 2 is where f prime is positive and f is increasing. The portion of the graph that moves downward in quadrant 3 then upward in quadrant 4 is where f prime is negative and f is decreasing. The portion of the graph that moves upward in quadrant 1 is where f prime is positive and f is increasing.
When we justify the properties of a function based on its derivative, we are using calculus-based reasoning.
Problem 1
These are two valid justifications for why a function f is an increasing function:
A. As the x-values increase, the values of f also increase.
B. The derivative of f is always positive.
Which of the above is a calculus-based justification?
Choose 1 answer:

Problem 2
The differentiable function f and its derivative f are graphed.
What is an appropriate calculus-based justification for the fact that f is decreasing when x>3?
Choose 1 answer:

Common mistake: Not relating the graph of the derivative and its sign.

When working with the graph of the derivative, it's important to remember that these two facts are equivalent:
  • f(x)<0 at a certain point or interval.
  • The graph of f is below the x-axis at that point/interval.
(The same goes for f(x)>0 and being above the x-axis.)

How f tells us where f has a relative minimum or maximum

In order for a function f to have a relative maximum at a certain point, it must increase before that point and decrease after that point.
At the maximum point itself, the function is neither increasing nor decreasing.
Function f is graphed. The x-axis is unnumbered. The graph consists of a curve. The curve starts in quadrant 3, moves upward, or is increasing, concave down to a relative max in quadrant 2, moves downward, or is decreasing, concave down until a point in quadrant 4 and then moves downward concave up to a point in quadrant 4, moves upward concave up, and ends in quadrant 1.
In the graph of the derivative f, this means that the graph crosses the x-axis at the point, so the graph is above the x-axis before the point and below the x-axis after.
Function f prime is graphed. The x-axis is unnumbered. The graph consists of U-shaped curve. The curve starts in quadrant 2, moves downward concave up through the negative x-axis, where f has a relative max, to a point in quadrant 4, moves upward concave up through a point on the positive x-axis, and ends in quadrant 1.
Problem 3
The differentiable function g and its derivative g are graphed.
What is an appropriate calculus-based justification for the fact that g has a relative minimum point at x=3?
Choose 1 answer:

Common mistake: Confusing the relationship between the function and its derivative

As we saw, the sign of the derivative corresponds to the direction of the function. However, we can't make any justification based on any other kinds of behavior.
For example, the fact that the derivative is increasing doesn't mean the function is increasing (or positive). Furthermore, the fact that the derivative has a relative maximum or minimum at a certain x-value doesn't mean the function must have a relative maximum or minimum at that x-value.
Problem 4
The differentiable function h and its derivative h are graphed.
Four students were asked to give an appropriate calculus-based justification for the fact that h is increasing when x>0.
Can you match the teacher's comments to the justifications?
1

Want more practice? Try this exercise.

Common mistake: Using obscure or non-specific language.

There are a lot of factors at play when we’re looking at the relationship between a function and its derivative: the function itself, that function’s derivative, the direction of the function, the sign of the derivative, etc. It's important to be extremely clear about what one is talking about at any given time.
For example, in Problem 4 above, the correct calculus-based justification for the fact that h increasing is that h is positive, or above the x-axis. One of the students' justifications was "It's above the x-axis." The justification didn't specify what is above the x-axis: the graph of h? The graph of h? Or maybe something else? Without being specific, such a justification cannot be accepted.

Want to join the conversation?

  • orange juice squid orange style avatar for user Steven Shelby
    is there any use to knowing the justification?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • eggleston blue style avatar for user dena escot
    I do not understand this statement "For example, the fact that the derivative is increasing doesn't mean the function is increasing (or positive)."
    (1 vote)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user elaw3670
      That other answer is incorrect. As per the quoted statement, there is no direct causation between the derivative increasing and either the function, itself, being positive OR increasing. Recall that the derivative is the slope at any given instant/point (ie, where the function is going): essentially that quote is stating that where the function is going is a separate matter from where the function is currently at. Say a continuous and differentiable function is approaching a local minimum between outputs f(x) = -1 and f(x) = -3; somewhere therein the function's derivative would be increasing (as the function as about to turn around -- ie: attain a derivative of 0,) while the function itself is both negative (below the x-axis,) AND decreasing (approaching the local minimum.)
      (3 votes)