Maxima and Minima

Introduction to Local Maxima and Local Minima

In many practical applications, we need to find the maximum or minimum values of a varying quantity. For a continuous function, the peaks and valleys on its graph represent the maximum and minimum values respectively within a certain neighborhood.

Local Maximum

A point $x=c$ is a point of local maximum if there is an interval $(c-h, c+h)$ around $c$ such that $f(c) \ge f(x)$ for all $x$ in that interval. The value $f(c)$ is called the local maximum value.

Local Minimum

A point $x=c$ is a point of local minimum if there is an interval $(c-h, c+h)$ around $c$ such that $f(c) \le f(x)$ for all $x$ in that interval. The value $f(c)$ is called the local minimum value.

Extreme Values

The local maximum and local minimum values of a function are collectively known as local extreme values.

These are called local because the comparison is made only with nearby points, not necessarily with all points in the entire domain.

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Critical Points and Turning Points

Before applying derivative tests, we need to identify candidate points where a maximum or minimum could occur.

Critical Point

A point $c$ in the domain of a function $f$ at which either $f'(c) = 0$ or $f$ is not differentiable is called a critical point of $f$.

Turning Point

A point where the graph of a continuous function changes its nature from strictly increasing to strictly decreasing or vice versa.

Important Note: Every point of local maximum or local minimum must be a critical point, but not every critical point gives a local extremum. Some critical points are neither maxima nor minima; for example, they may be stationary points where the curve continues to increase on both sides, or points of inflection.

First Derivative Test

Let $f$ be a function defined on an open interval $I$. Let $f$ be continuous at a critical point $c$ in $I$. Then:

1.Local Maximum

If $f'(x)$ changes sign from positive to negative as $x$ increases through $c$ (that is, $f'(x) > 0$ for $x$ slightly less than $c$, and $f'(x) < 0$ for $x$ slightly greater than $c$), then $c$ is a point of local maximum.

2.Local Minimum

If $f'(x)$ changes sign from negative to positive as $x$ increases through $c$ (that is, $f'(x) < 0$ for $x < c$, and $f'(x) > 0$ for $x > c$), then $c$ is a point of local minimum.

3.Neither Maximum nor Minimum

If $f'(x)$ does not change sign as $x$ increases through $c$, then $c$ is neither a point of local maximum nor a point of local minimum.

Sometimes such a point may be a point of inflection, but that requires an additional change in concavity; it is not automatic merely because $f'(x)$ does not change sign.

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Second Derivative Test

This test is often quicker than the First Derivative Test when the second derivative is easy to compute.

Let $f$ be a function defined on an interval $I$ and let $c \in I$. Suppose $f$ is twice differentiable at $c$. Then:

1.Local Maximum

$x=c$ is a point of local maximum if $f'(c) = 0$ and $f''(c) < 0$. In this case, the curve is concave downward at $c$.

2.Local Minimum

$x=c$ is a point of local minimum if $f'(c) = 0$ and $f''(c) > 0$. In this case, the curve is concave upward at $c$.

3. Test Fails

If $f'(c) = 0$ and $f''(c) = 0$, then the test is inconclusive. In that case, we must use the First Derivative Test or some other method.

So, the second derivative test is a shortcut, but it does not always decide the nature of a critical point.

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Solved Examples

Example 1: Local Maxima and Minima of a Polynomial

Find all points of local maxima and local minima of the function $f(x) = x^3 - 3x + 3$.

Solution: Step 1: Find the first derivative $f'(x)$. $$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1).$$

Step 2: Find the critical points by setting $f'(x) = 0$. $$3(x - 1)(x + 1) = 0 \implies x = 1,\; x = -1.$$ So the possible points of local extrema are $x=1$ and $x=-1$.

Step 3: Find the second derivative. $$f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x.$$

Step 4: Test each critical point.

• At $x=1$: $$f''(1)=6>0.$$ Therefore the graph is concave upward there, so $x=1$ is a point of local minimum. The local minimum value is $$f(1)=1^3-3(1)+3=1.$$
• At $x=-1$: $$f''(-1)=-6<0.$$ Therefore the graph is concave downward there, so $x=-1$ is a point of local maximum. The local maximum value is $$f(-1)=(-1)^3-3(-1)+3=-1+3+3=5.$$

Answer: Local maximum at $x=-1$ with value $5$; local minimum at $x=1$ with value $1$.

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Example 2: Critical Point Which Is Not an Extremum

Find all points of local maxima and local minima of the function $f(x) = 2x^3 - 6x^2 + 6x + 5$.

Solution: Step 1: Find the first derivative. $$f'(x) = 6x^2 - 12x + 6 = 6(x^2 - 2x + 1) = 6(x - 1)^2.$$

Step 2: Find the critical points. $$6(x-1)^2=0 \implies x=1.$$ So the only critical point is $x=1$.

Step 3: Try the second derivative test. $$f''(x)=12(x-1).$$ At $x=1$, $$f''(1)=0.$$ So the second derivative test fails.

Step 4: Use the first derivative test. Observe that $$f'(x)=6(x-1)^2 \ge 0$$ for all real $x$, and in fact it is positive for all $x\ne 1$. Thus:

• for $x<1$, $f'(x)>0$,
• for $x>1$, $f'(x)>0$.

Step 5: Conclude. Since the derivative does not change sign at $x=1$, the function is increasing on both sides of $x=1$. Therefore, $x=1$ is neither a point of local maximum nor a point of local minimum. In this case, it is a stationary point of inflection.

Answer: The function has no local maximum and no local minimum. The critical point $x=1$ is a stationary point of inflection.

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Example 3: A Function with No Maxima or Minima

Find the local maxima and local minima, if any, for the function $f(x) = e^x$.

Solution: Step 1: Differentiate the function. $$f'(x)=e^x.$$

Step 2: Find critical points by solving $f'(x)=0$. That gives $$e^x=0.$$ But this equation has no real solution because $e^x>0$ for every real number $x$.

Step 3: Conclude. Since the derivative is never zero and is never undefined, the function has no critical points. Without any critical points, the function cannot have a local maximum or a local minimum. Also, because $f'(x)>0$ for all $x$, the function is strictly increasing everywhere.

Answer: The function has no local maxima and no local minima.

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Example 4: Rational Function Maxima and Minima

Find the local maximum and local minimum values of the function $f(x) = x + \frac{1}{x}$, $x \ne 0$.

Solution: Step 1: Differentiate the function. $$f'(x)=1-\frac{1}{x^2}.$$

Step 2: Find critical points. Set $f'(x)=0$: $$1-\frac{1}{x^2}=0 \implies 1=\frac{1}{x^2} \implies x^2=1 \implies x=1, -1.$$

Step 3: Find the second derivative. $$f''(x)=\frac{d}{dx}\left(1-x^{-2}\right)=2x^{-3}=\frac{2}{x^3}.$$

Step 4: Test each critical point.

• At $x=1$: $$f''(1)=2>0,$$ so $x=1$ is a point of local minimum. The local minimum value is $$f(1)=1+1=2.$$
• At $x=-1$: $$f''(-1)=-2<0,$$ so $x=-1$ is a point of local maximum. The local maximum value is $$f(-1)=-1-1=-2.$$

Because the function is defined on two separate intervals $(-\infty,0)$ and $(0,\infty)$, it is completely possible that the local maximum value $-2$ is numerically less than the local minimum value $2$.

Answer: Local maximum value is $-2$ at $x=-1$. Local minimum value is $2$ at $x=1$.

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Example 5: Trigonometric Function

Find the local maximum and minimum values of $f(x) = \sin x - \cos x$, where $0 < x < 2\pi$.

Solution: Step 1: Differentiate the function. $$f'(x)=\cos x+\sin x.$$

Step 2: Find critical points by setting $f'(x)=0$. $$\cos x+\sin x=0 \implies \sin x=-\cos x \implies \tan x=-1.$$ In the interval $(0,2\pi)$, this happens at $$x=\frac{3\pi}{4}, \quad x=\frac{7\pi}{4}.$$

Step 3: Find the second derivative. $$f''(x)=-\sin x+\cos x.$$

Step 4: Test the critical points.

• At $x=\frac{3\pi}{4}$: $$f''\left(\frac{3\pi}{4}\right)=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}<0.$$ So this is a local maximum. The maximum value is $$f\left(\frac{3\pi}{4}\right)=\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)=\sqrt{2}.$$
• At $x=\frac{7\pi}{4}$: $$f''\left(\frac{7\pi}{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0.$$ So this is a local minimum. The minimum value is $$f\left(\frac{7\pi}{4}\right)=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}.$$

Answer: Local maximum value is $\sqrt{2}$ at $x=\frac{3\pi}{4}$; local minimum value is $-\sqrt{2}$ at $x=\frac{7\pi}{4}$.

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Example 6: First Derivative Test Where Second Derivative Test Fails

Find local maxima and minima for $f(x) = x^4$.

Solution: Step 1: Find the first derivative. $$f'(x)=4x^3.$$

Step 2: Find the critical points. $$4x^3=0 \implies x=0.$$

Step 3: Try the second derivative test. $$f''(x)=12x^2.$$ At $x=0$, $$f''(0)=0.$$ So the second derivative test is inconclusive.

Step 4: Use the first derivative test.

• For $x<0$, say $x=-0.1$, $$f'(x)=4x^3<0.$$
• For $x>0$, say $x=0.1$, $$f'(x)=4x^3>0.$$ The derivative changes from negative to positive as we pass through 0.

Step 5: Conclude. Therefore $x=0$ is a point of local minimum. The local minimum value is $$f(0)=0.$$ There is no local maximum.

Answer: Local minimum at $x=0$ with value $0$.

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Example 7: Non-Differentiable Critical Point

Find all points of local maxima and local minima of the function $f(x) = |x|$.

Solution: Step 1: Note that $f(x)=|x|$ is not differentiable at $x=0$. Hence $x=0$ is a critical point.

Step 2: Since the derivative does not exist at $x=0$, derivative tests are not directly applicable there. So we use the basic definition of local extrema.

Step 3: Compare nearby values. $$f(0)=|0|=0.$$ For every $x\ne 0$, $$f(x)=|x|>0.$$ Thus, in every neighborhood of 0, $$f(x) \ge f(0).$$

Step 4: Conclude. So $x=0$ is a point of local minimum, and the local minimum value is 0. There is no local maximum because the function increases away from 0 on both sides.

Answer: Local minimum at $x=0$ with value $0$; no local maximum.

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Example 8: Finding Maximum Profit

A company's profit function is given by $P(x) = 41 - 24x - 18x^2$. Find the maximum profit that the company can make.

Solution: Step 1: Differentiate the profit function. $$P'(x)=-24-36x.$$

Step 2: Find the critical point. $$-24-36x=0 \implies -36x=24 \implies x=-\frac{2}{3}.$$

Step 3: Check the second derivative. $$P''(x)=-36<0.$$ So mathematically the parabola opens downward and attains its greatest value at $$x=-\frac{2}{3}.$$ The corresponding value is $$P\left(-\frac{2}{3}\right)=41-24\left(-\frac{2}{3}\right)-18\left(-\frac{2}{3}\right)^2=41+16-8=49.$$

Step 4: Interpret the practical meaning. In a real profit problem, $x$ usually represents number of units, so negative values of $x$ are not meaningful. Therefore, for the practical domain $x \ge 0$, the function is decreasing because $$P'(x)=-24-36x<0 \quad \text{for all } x\ge 0.$$ Hence the maximum practical profit occurs at the left endpoint $x=0$: $$P(0)=41.$$

Answer: Mathematically, the maximum value of the function is $49$ at $x=-\frac{2}{3}$. Practically, if $x$ denotes number of units produced and hence $x \ge 0$, the maximum profit is $41$ at $x=0$.