How the Second Derivative Describes Graph Shape
The first derivative describes how a function’s slope changes from point to point. The second derivative describes how that slope is changing. This gives the second derivative a geometric role: its sign tells us whether slopes are increasing or decreasing, and therefore whether the graph bends upward or downward.
The sign must be interpreted over an interval or neighborhood, not just read at one isolated input. A positive second derivative at a point does not automatically tell us the concavity on either side unless we know enough about nearby values.
Concavity and the Second Derivative
For a function \(f\) that is twice differentiable on an interval, the second derivative is the derivative of the first derivative: \(f''(x)=(f'(x))'\). In particular, it measures how \(f'(x)\), the slope of the graph, changes as \(x\) changes.
Visually, a concave-up graph bends like a cup, and a concave-down graph bends like an upside-down cup. These descriptions refer to the direction of bending, not whether the function is positive or negative. A graph can lie below the \(x\)-axis and still be concave up.
For a twice-differentiable function, the connection with slope follows from the Mean Value Theorem applied to \(f'\): if \(f''>0\) on an interval, then \(f'\) increases there; if \(f''<0\), then \(f'\) decreases. The next result makes the geometric consequence more precise.
Why Concavity Determines the Position Relative to a Tangent
The tangent line at \(x=a\) is the line through \((a,f(a))\) with slope \(f'(a)\). Its equation is \(y=f(a)+f'(a)(x-a)\). To compare the graph with this line, compare the change in \(f\) from \(a\) to \(x\) with the change predicted by the slope at \(a\).
If \(f''>0\) throughout the interval, then \(f'\) is strictly increasing. Thus, when \(x>a\), any slope \(f'(c)\) with \(a<c<x\) is greater than \(f'(a)\).
For \(x>a\), some \(c\) between \(a\) and \(x\) satisfies \(f(x)-f(a)=f'(c)(x-a)\). Since \(f'(c)>f'(a)\) and \(x-a>0\), this change is greater than \(f'(a)(x-a)\). Hence \(f(x)>f(a)+f'(a)(x-a)\).
For \(x<a\), the point \(c\) between \(x\) and \(a\) has \(f'(c)<f'(a)\). Now \(x-a<0\), so multiplying reverses the comparison: \(f'(c)(x-a)>f'(a)(x-a)\). The same tangent-line inequality follows. If \(f''<0\), the slope comparisons reverse, giving the graph below the tangent instead.
This argument gives a reason for the cup-shaped picture: on a concave-up interval, slopes to the right of \(a\) are larger than the tangent slope, while slopes to the left are smaller. The graph remains above the tangent on both sides.
Worked Example: Positive Second Derivative
Worked Example: Positive Second Derivative
Let \(f(x)=x^2+1\). Determine its concavity and compare its graph with the tangent line at \(x=1\).
First differentiate:
Since \(f''(x)=2>0\) for every real \(x\), the function is concave up everywhere. At \(x=1\), \(f(1)=1^2+1=2\) and \(f'(1)=2(1)=2\). The tangent line is therefore
The difference between the function and the tangent line is
For \(x\ne1\), \((x-1)^2>0\), so \(f(x)>2x\): the graph lies above the tangent line, as the theorem predicts. At \(x=1\), both sides equal \(2\). Differentiating the tangent equation gives slope \(2\), which agrees with \(f'(1)=2\); the tangent and the graph meet with the required slope.
Worked Example: Negative Second Derivative
Worked Example: Negative Second Derivative
Let \(g(x)=-x^2+4x\). Determine its concavity and compare it with the tangent line at \(x=1\).
Differentiate twice:
Because \(g''(x)=-2<0\) for every real \(x\), \(g\) is concave down everywhere. At \(x=1\), \(g(1)=-(1)^2+4(1)=3\), and \(g'(1)=-2(1)+4=2\). Thus the tangent line is
Subtract the tangent expression from the function:
For \(x\ne1\), \(-(x-1)^2<0\), so \(g(x)<2x+1\). The graph lies below its tangent line, as expected for a concave-down function. At \(x=1\), the difference is zero, and the tangent line’s slope \(2\) agrees with \(g'(1)=2\).
What Does a Zero Second Derivative Mean?
If \(f''(a)=0\), then the second derivative has no positive or negative value at that one input. This fact alone does not determine the concavity on either side of \(a\), and it does not guarantee that concavity changes there. It is necessary to inspect the sign of \(f''\) on intervals to the left and right.
When \(f''\) is continuous at \(a\), a strictly positive value \(f''(a)>0\) ensures that it remains positive sufficiently close to \(a\); likewise, a strictly negative value ensures it remains negative nearby. But \(f''(a)=0\) gives no such sign information. The sign test must be carried out on each side.
Worked Example: Zero Second Derivative With No Change in Concavity
Worked Example: Zero Second Derivative With No Change in Concavity
Consider \(p(x)=x^4\) at \(x=0\). Compute its second derivative and test whether the concavity changes there.
Differentiate:
At \(x=0\), \(p''(0)=12(0)^2=0\). But for every \(x\ne0\), \(12x^2>0\), both to the left and to the right of zero. Thus \(p\) is concave up on both sides; its concavity does not change at zero. The graph and its horizontal tangent line \(y=0\) meet at the origin, and \(p(x)-0=x^4>0\) for \(x\ne0\). Therefore zero second derivative does not guarantee an inflection point.
Worked Example: Zero Second Derivative With a Change in Concavity
Worked Example: Zero Second Derivative With a Change in Concavity
Now consider \(q(x)=x^3\) at \(x=0\). Determine whether the concavity changes there.
Differentiate twice:
At zero, \(q''(0)=6(0)=0\). For \(x<0\), \(6x<0\), so the function is concave down. For \(x>0\), \(6x>0\), so it is concave up. Since concavity changes across zero, the origin is an inflection point. The function is continuous there, with \(q(0)=0\). This example shows that a zero second derivative can occur at an inflection point, but the sign change—not the zero by itself—establishes the change in concavity.
A Reliable Sign-Checking Method
When asked to interpret concavity near a point, organize the reasoning around intervals rather than a single substitution.
Differentiate twice and simplify \(f''(x)\).
Find where \(f''(x)=0\) or where \(f''\) is undefined; these values divide the domain into intervals to check.
Use a test value or a sign analysis to determine whether \(f''\) is positive or negative throughout each interval.
Positive \(f''\) means concave up; negative \(f''\) means concave down. A change of sign across a point indicates a change in concavity there, provided the graph has the point in question.
Common Mistake: Treating \(f''(a)=0\) as a Conclusion
A frequent error is to find a zero of the second derivative and immediately label it an inflection point. For \(p(x)=x^4\), the incorrect calculation would be: “\(p''(0)=12(0)^2=0\), so \(x=0\) is an inflection point.” The calculation of \(p''(0)\) is correct, but the conclusion does not follow. Since \(p''(x)=12x^2>0\) on both sides of zero, concavity does not change there.
The correction is to check the sign on both sides. For \(q(x)=x^3\), \(q''(x)=6x\) is negative to the left and positive to the right, so its concavity does change at zero. The two examples show why the sign test, not the equation \(f''(a)=0\) alone, is decisive.
Check Your Understanding
Use the sign of the second derivative and, where relevant, compare the graph with a tangent line.
- If \(r''(x)>0\) throughout an interval, what happens to the slopes of \(r\) as \(x\) increases, and how is the graph positioned relative to a tangent line on that interval?
- For \(s(x)=-3x^2\), find \(s''(x)\) and state the concavity.
- For \(t(x)=x^5\), find \(t''(x)\). Does its sign change at zero?
- Why does \(f''(a)=0\) fail, by itself, to establish an inflection point?
- What sign pattern in \(f''\) on the two sides of a point indicates a change in concavity?
