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Dated: 30-10-2024

Extreme Valued Theorem

Theorem

If a function1 is continuous2 over the closed interval3 \([a, b]\) then the function1 has absolute maximum4 and absolute minimum4 values over that interval.3

Steps

If \(f(x)\) is continuous2 over \([a, b]\) then

  1. Find critical points4 and their corresponding \(f(x)\) values.
  2. Find \(f(a)\) and \(f(b)\).
  3. Largest among all these values is absolute maximum and smallest is absolute minimum.

Example

\[f(x) = x^3 + x^2 - x + 1 \text { on } \left[-2 , \frac 1 2\right]\]
\[f^{\prime}(x) = 3x^2 + 2x - 1\]
\[\because \text{For critical points } f^{\prime}(x) = 0\]
\[0 = 3x^2 + 2x - 1\]

Simplify and we get

\[x = -1 , \frac 1 3\]

Putting all these values in \(f(x)\), we get

\[f(-2) = (-2)^3 + (-2)^2 - (-2) + 1 = -1\]
\[f(-1) = (-1)^3 + (-1)^2 - (-1) + 1 = 2\]
\[f\left(\frac 1 3\right) = \left(\frac 1 3\right)^3 + \left(\frac 1 3 \right)^2 - \left(\frac 1 3 \right) + 1 = \frac{22}{27}\]
\[f\left(\frac 1 2\right) = \left(\frac 1 2\right)^3 + \left(\frac 1 2 \right)^2 - \left(\frac 1 2 \right) + 1 = \frac{7}{8}\]

This shows that absolute maximum is \(2\) present at \(x = -1\) and absolute minimum is \(-1\) present at \(x = -2\).

Steps for function1 of 2 Variables

  1. Find critical points4 of \(f\) that lie in the interior of \(R\).
  2. Find all boundary points where absolute extrema can occur.
  3. Evaluate \(f(x, y)\) at these points. The largest of them is absolute maximum and the smallest being absolute minimum.

Example

Find absolute minimum and absolute maximum values over first quadrant5 within the triangular region \(R\) bounded by \(x = 0\), \(y = 0\) and \(y = 9 - x\) where function1 is

\[f(x, y) = 2 + 2x + 2y - x^2 - y^2\]

Solution

From the equation \(y = 9 - x\), we get 3 corners, \((0, 0)\), \((9, 0)\) and \((0, 9)\). To find other points \(\(f_x(x, y) = 2 - 2x\)\) \(\(f_y(x, y) = 2 - 2y\)\) Assuming \(f_x = 0\) and \(f_y = 0\), we get \((1, 1)\).

Now we try to look at the sides of the triangle
For the x axis, we have \(f(x, 0) = 2 + 2x - x^2\).

\[f^{\prime}(x, 0) = 2 - 2x\]
\[\because f^{\prime}(x, 0) = 0\]
\[0 = 2 - 2x\]

This gives us \((1, 0)\)
Similarly, \(f^{\prime}(0, y)\) will give us \((0, 1)\)

For the side \(y = 9 - x\),

\[f(x, 9 - x) = 2 + 2x + 2(9 - x) -x^2 - (9 - x)^2\]
\[f^{\prime}(x, 9 - x) = 18 - 4x\]

Therefore, we have \(x = \frac 9 2\)
Similarly, we will get \(y = \frac 9 2\)

\((x, y)\) \(f(x, y)\)
\((0, 0)\) \(2\)
\((9, 0)\) \(-61\)
\((1, 0)\) \(3\)
\((\frac 9 2, \frac 9 2)\) \(\frac {-41}{2}\)
\((0, 9)\) \(-61\)
\((0, 1)\) \(3\)
\((1, 1)\) \(4\)

Therefore, the absolute minimum is \(-61\) existing at \((0, 9)\) and \((9, 0)\) and absolute maximum is \(4\) existing at \((1, 1)\).

References

Read more about notations and symbols.

  1. Read more about functions. ↩↩↩↩

  2. Read more about continuity. ↩↩

  3. Read more about intervals. ↩↩

  4. Read more about extreme values. ↩↩↩↩

  5. Read more about quadrants. ↩