Introduction
A tangent line describes a curve near a point, but concavity gives it a stronger role: across an interval where a function is concave in one direction, the tangent line at any point bounds the function throughout that interval. For a concave-up function, the tangent line lies below the graph. For a concave-down function, it lies above.
The word “throughout” matters. This is not a claim about how well a line approximates a curve only very close to the point of tangency. It is a global inequality on the interval where the function has the stated concavity. The interval and differentiability conditions are therefore part of the result.
Concavity and Tangent Lines
Let \(I\) be an interval and let \(f:I\to\mathbb{R}\). Concavity compares the value of the function between two points with the values at those points. In the convention used here, a concave-up function satisfies the inequality below; a concave-down function satisfies the reverse inequality.
For a differentiable function, concavity up on an interval is equivalent to \(f'\) being nondecreasing there; concavity down is equivalent to \(f'\) being nonincreasing. This is the derivative characterization of concavity. It lets us prove the tangent-line result by comparing derivative values on the interval between the tangency point and the point being considered.
- If \(f\) is concave up on \(I\), then for every \(a,x\in I\), $$ f(x)\geq f(a)+f'(a)(x-a). $$
- If \(f\) is concave down on \(I\), then for every \(a,x\in I\), $$ f(x)\leq f(a)+f'(a)(x-a). $$
At \(x=a\), both sides are equal. At other points the inequality may also be an equality: a function can be linear on part of the interval, for example. Concavity guarantees the direction of the bound, not necessarily a strict gap.
Why the Tangent-Line Inequality Holds
We prove the concave-up case. Fix \(a,x\in I\). If \(x\neq a\), the Mean Value Theorem applies to the segment with endpoints \(a\) and \(x\): differentiability on \(I\) ensures continuity on that segment and differentiability between its endpoints. It gives a point \(c\) strictly between \(a\) and \(x\) such that
Since \(f\) is concave up, \(f'\) is nondecreasing. The comparison between \(f'(c)\) and \(f'(a)\) depends on which side of \(a\) contains \(x\). Keeping track of that sign is essential.
Then \(a<c<x\), so \(f'(c)\geq f'(a)\). Because \(x-a>0\), multiplication preserves the inequality: $$ f'(c)(x-a)\geq f'(a)(x-a). $$
Then \(x<c<a\), so \(f'(c)\leq f'(a)\). Because \(x-a<0\), multiplication reverses the inequality: $$ f'(c)(x-a)\geq f'(a)(x-a). $$
In either case, $$ f(x)-f(a)\geq f'(a)(x-a). $$ Adding \(f(a)\) gives \(f(x)\geq f(a)+f'(a)(x-a)\). If \(x=a\), this is equality directly.
This proves the theorem for concave-up functions. For a concave-down function, \(f'\) is nonincreasing. The derivative comparisons in both cases reverse, and the sign of \(x-a\) still determines whether multiplication reverses an inequality. The result is \(f'(c)(x-a)\leq f'(a)(x-a)\), which gives the stated upper bound.
Worked Examples
Worked Example: A Concave-Up Parabola
Find the tangent line to \(f(x)=x^2\) at \(a=2\), and verify that the function lies above it for every real \(x\).
First calculate the function value and derivative at the tangency point:
The tangent line is therefore $$ y=f(2)+f'(2)(x-2)=4+4(x-2)=4x-4. $$
To check its position, subtract the tangent-line value from the function value:
Thus \(x^2\geq 4x-4\) for every real \(x\), with equality at \(x=2\). This agrees with the theorem because \(f'(x)=2x\) is increasing on \(\mathbb{R}\), so \(f\) is concave up there.
As a direct check at the tangency point, the line gives \(4(2)-4=4=f(2)\).
Worked Example: A Concave-Down Parabola
Find the tangent line to \(f(x)=-x^2+4x\) at \(a=1\), and check that the graph lies below the line.
The function value and derivative at \(1\) are
The tangent line is $$ y=3+2(x-1)=2x+1. $$
Subtract the line from the function:
So \(f(x)\leq 2x+1\) for every real \(x\), with equality at \(x=1\). The derivative \(f'(x)=-2x+4\) is decreasing, confirming that \(f\) is concave down. Substitution into the tangent line gives \(2(1)+1=3=f(1)\), as required.
Worked Example: Using a Tangent as a Bound on a Restricted Interval
Let \(f(x)=x^3\) on \([0,2]\). Find its tangent line at \(a=1\), then use it to bound \(f(0)\) and \(f(2)\).
On this interval, \(f'(x)=3x^2\) is nondecreasing: if \(0\leq u\leq v\leq 2\), then \(u^2\leq v^2\), so \(3u^2\leq 3v^2\). Thus \(f\) is concave up on \([0,2]\). At the tangency point,
The tangent line is $$ y=1+3(x-1)=3x-2. $$
The theorem gives \(x^3\geq 3x-2\) for every \(x\in[0,2]\). To verify the inequality directly, factor the difference:
For \(x\in[0,2]\), both factors are nonnegative. At \(x=0\), the line gives \(-2\), and \(f(0)=0\geq -2\). At \(x=2\), the line gives \(4\), and \(f(2)=8\geq 4\). At the tangency point, the line gives \(3(1)-2=1=f(1)\). These checks agree with the theorem and the factorization.
A Common Sign Error
The proof for points to the left of \(a\) is a common source of mistakes. When \(x<a\), the factor \(x-a\) is negative. Multiplying by it reverses an inequality; treating it as positive can produce the wrong tangent-line position.
Worked Example: Correcting the Left-Side Sign
Consider \(f(x)=x^2\), with tangent point \(a=2\), and test \(x=0\). The Mean Value Theorem gives a point \(c\) between \(0\) and \(2\) such that
Since \(f(0)-f(2)=0-4=-4\) and \(f'(c)=2c\), this equation requires \(-4=2c(-2)=-4c\), so \(c=1\). Now \(f'(c)=2\leq 4=f'(2)\). The incorrect step would be to preserve the inequality after multiplying by \(-2\), claiming \(-4\leq -8\). That is false.
The correct multiplication reverses the inequality: $$ 2\leq 4 \quad\Longrightarrow\quad 2(-2)\geq 4(-2), $$ so \(-4\geq -8\). Indeed, \(f(0)-f(2)=-4\), while the tangent-line increment is \(f'(2)(0-2)=4(-2)=-8\). Equivalently, \(f(0)=0\) lies above the tangent-line value \(4(0)-4=-4\).
What the Result Does and Does Not Say
The theorem provides a one-sided bound on the entire interval of concavity. For a concave-up function, the tangent line is a lower bound; for a concave-down function, it is an upper bound. This can turn a geometric observation into an algebraic inequality, as the examples show.
The interval condition cannot be discarded. If a function changes concavity, a tangent line established as a lower bound on one interval need not remain a lower bound beyond it. Likewise, the result requires a tangent at \(a\), so the derivative \(f'(a)\) must exist. A concavity statement alone does not supply a tangent at a point where the function is not differentiable.
Finally, “lies above” and “lies below” allow equality. At the point of tangency equality always holds, and a linear portion of the graph can produce equality at additional points. The conclusion is a weak inequality, not a guarantee that the graph stays strictly separated from its tangent.
Check Your Understanding
Use the concavity and tangent-line ideas from this tutorial to answer the following questions.
- State the tangent-line inequality for a differentiable concave-up function on an interval.
- Why does the Mean Value Theorem proof require separate attention when \(x<a\)?
- For \(f(x)=x^2\) at \(a=1\), find the tangent line and verify algebraically that it is a lower bound on \(\mathbb{R}\).
- If a differentiable function is concave down on an interval, is its tangent line a lower bound or an upper bound there?
- Can equality hold at a point other than the point of tangency? Explain briefly.
