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# Justification using second derivative

"Calculus-based reasoning" with the second derivative of a function can be used to justify claims about the concavity of the original function and about its inflection points.
We've learned that the first derivative f, prime gives us information about where the original function f increases or decreases, and about where f has extremum points.
The second derivative f, start superscript, prime, prime, end superscript gives us information about the concavity of the original function f and about where f has inflection points.

## Let's review what concavity is all about.

A function is concave up when its slope is increasing. Graphically, a graph that's concave up has a cup shape, \cup.
Function f is graphed. The x-axis is unnumbered. The graph consists of a curve. The curve starts in quadrant 2, moves downward concave up to the y-axis, moves upward concave up through 2 points, and ends in quadrant 1. Tangent lines move upward and touch each of the 2 points. The line tangent to the higher point, at a higher x-value, has a steeper slope than the line tangent to the lower point.
The graph of f is concave up (notice its \cup shape). Notice how, as x increases, the slope is increasing.
Similarly, a function is concave down when its slope is decreasing. Graphically, a graph that's concave down has a cap shape, \cap.
Function g is graphed. The x-axis is unnumbered. The graph consists of a curve. The curve starts in quadrant 2, moves upward concave down to the y-axis, moves downward concave down through 2 points, and ends in quadrant 4. Tangent lines move downward and touch each of the 2 points. The line tangent to the lower point, at a higher x-value, has a steeper slope than the line tangent to the higher point.
The graph of g is concave down (notice its \cap shape). Notice how, as x increases, the slope is decreasing.
A point of inflection is where a function changes concavity.
Function f is graphed. The x-axis is unnumbered. The graph consists of a curve. The curve starts in quadrant 2, moves downward concave up to a point in quadrant 1, moves upward concave up to an inflection point, continues upward concave down to a point, moves downward concave down and ends in quadrant 4.

## How $f''$f, start superscript, prime, prime, end superscript informs us about the concavity of $f$f

When the second derivative f, start superscript, prime, prime, end superscript is positive, that means the first derivative f, prime is increasing, which means that f is concave up. Similarly, a negative f, start superscript, prime, prime, end superscript means f, prime is decreasing and f is concave down.
f, start superscript, prime, prime, end superscriptf, primef
positive plusincreasing \nearrowconcave up \cup
negative minusdecreasing \searrowconcave down \cap
crossing x-axis (changes sign)extremum point (changes direction)inflection point (changes concavity)
Here’s a graphical example:
f, start superscript, prime, prime, end superscriptf, primef
Notice how f is start color #aa87ff, start text, c, o, n, c, a, v, e, space, d, o, w, n, end text, end color #aa87ff to the left of x, equals, c and start color #1fab54, start text, c, o, n, c, a, v, e, space, u, p, end text, end color #1fab54 to the right of x, equals, c.
Problem 1
Let f be a twice differentiable function. This is the graph of its second derivative, f, start superscript, prime, prime, end superscript.
Over which interval is f always concave up?

### Common mistake: Confusing the relationship between $f$f, $f'$f, prime, and $f''$f, start superscript, prime, prime, end superscript

Remember that for f to be concave up, f, prime needs to be increasing and f, start superscript, prime, prime, end superscript needs to be positive. Other behaviors of f, f, prime, and f, start superscript, prime, prime, end superscript aren't necessarily related.
For example, in Problem 1 above, f, start superscript, prime, prime, end superscript is concave up over the interval open bracket, minus, 8, comma, minus, 2, close bracket but it doesn't mean f is concave up on that interval.
Problem 2
Let h be a twice differentiable function. This is the graph of its second derivative, h, start superscript, prime, prime, end superscript.
Where does h have an inflection point?

Want more practice? Try this exercise.

### Common mistake: Misinterpreting the graphical information presented

Imagine a student solving Problem 2 above, thinking that the graph is of the first derivative of h. In that case, h would have an inflection point at A and B, because these are the points where h, prime changes its direction. This student would be wrong, because this is the graph of the second derivative and the correct answer is D.
Remember to always make sure you understand the information given. Are we given the graph of the function f, the first derivative f, prime, or the second derivative f, start superscript, prime, prime, end superscript?
Problem 3
The twice differentiable function g and its second derivative g, start superscript, prime, prime, end superscript are graphed.
Four students were asked to give an appropriate calculus-based justification for the fact that g has an inflection point at x, equals, minus, 2.
Can you match the teacher's comments to the justifications?

## Using the second derivative to determine whether an extremum point is a min or a max

Imagine we are given that a function f has an extremum point at x, equals, 1, and that it's concave up over the interval open bracket, 0, comma, 2, close bracket. Can we tell, based on this information, whether that extremum point is a minimum or a maximum?
The answer is YES. Recall that a function that's concave up has a cup \cup shape. In that shape, a curve can only have a minimum point.
Similarly, if a function is concave down when it has an extremum, that extremum must be a maximum point.
Function f is graphed. The x-axis is unnumbered. The graph consists of a curve. The curve starts in quadrant 2, moves downward concave up to a minimum point in quadrant 1, moves upward concave up and then concave down to a maximum point in quadrant 1, moves downward concave down and ends in quadrant 4.
Problem 4
The twice differentiable function h and its second derivative h, start superscript, prime, prime, end superscript are graphed.
Given that h, prime, left parenthesis, minus, 4, right parenthesis, equals, 0, what is an appropriate calculus-based justification for the fact that h has a relative maximum at x, equals, minus, 4?

Want more practice? Try this exercise.

## Want to join the conversation?

• If the first derivative of a function at c is zero, does that mean that the second derivative at c is also zero? • Not necessarily. Take x^2. First derivative at 0 is 2*0, which is 0, but its second derivative is just a constant 2, so at x=0 the constant equation 2 is 2 everywhere.

Another way to look at it is the first derivative tells if the slope is 0, and the second derivative will tell if the original function is at an inflection point. If the slope of a function is 0, does that necessarily mean that it is also an inflection point? Any relative extrema disprove this. Let me know if that didn't make sense.
• The graph under the title "How f′′ informs us about the concavity of f" is wrong.
Slope of f at point c should be zero, but it's clearly not the case in the graph.
Correct graph should look something like y=x^3, with appropriate shifts.
(1 vote) • No, it really isn't wrong.

The slope at point c does NOT need to be zero. It's the second derivative (the slope of the slope as it were) that is zero at an inflection point (a change of concavity).
It is true that y = x³ has an inflection point at x = 0, and that the slope at x = 0 is also 0, but this is just coincidence. It's that fact that f''(0) = 6x = 0 that indicates a change in concavity.

Consider the sine curve. To the left of x = 0 it is clearly "concave up" - it's a U shape. To the right it's clearly concave down a sort of ∩ shape. the change occurs at x = 0, even though the slope at x = 0 is obviously greater than zero. (cos(0) = 1). However the second derivative (-sin(0)) is zero at that point.
• I have a quick clarification. Just knowing that f′′(x) (the second derivative of f(x)) is less than zero, does that tell me right there without knowing anything else that there is a maximum point. That's what I was confused at, because that's what I thought Sal was saying, but that can't be right, because there are lot's of times that second derivatives of functions are negative, but it's not a relative maximum.

Isn't this what's right: If f′′(x) < 0 && f′(x) === 0 then you have a relative maximum? Or if you have f′′(x) > 0 && f′(x) === 0 you have a relative minimum?

- ncochran2 • There is a relative maximum if:
1. f''(x) < 0 and f'(x) = 0 if the function is twice-differentiable at x
Example: -x^2, x = 0
2. f'(x) changes from positive to negative around x, and f'(x) = 0 or undefined, for f once-differentiable around x (really the same thing, just if you have a cusp)
Example: -|x|, x = 0
3. f has a point discontinuity at x, but the point is a relative maximum
Example: -x^2 if x ≠ 0, 1 if x = 0
4. Miscellaneous things: jump discontinuities, everywhere continuous but nowhere differentiable functions, almost continuous functions, nowhere continuous functions, etc. Don't worry about these.
Example: Weierstrass function

So yeah. Don't forget about if f' is undefined.  