MATH 121 (Wg Cdr Monir)
Maxima and Minima
♦ Increasing Function:
If \(y = f(x)\) is a function of x and y increases as x increases in a certain interval, Then y is called an increasing function of x in that interval.
\begin{align*} \frac{dy}{dx} = & \tan \Psi >0 \\ & \Psi < 90^\circ \end{align*}
♦ Decreasing Function:
If \(y = f(x)\) is a function of x and y decreases as x increases in a certain interval, then y is called a decreasing function of x in that interval.
\begin{align*} \frac{dy}{dx} = & \tan \Psi <0 \\ & \Psi > 90^\circ \end{align*}
♦ Concavity:
If the graph of a function \(f\) lies above all of its tangents on an interval l, then it is said to be concave up on l.
If the graph of a function \(f\) lies below all of its tangent on an interval l, then it is said to be concave down.
♦ Test for concavity
• The function \(f\), is concave up on any open interval l where \(f''(x)>0\) and it is concave down where \(f''(x)<0\)
• Inflection Point: If \(x = a\) inflection point, when
(1) \(f''(x)>0\) when \(x<a\) and \(f''(x)<0\) when \(x>a\)
or
(2) \(f''(x)<0\) when \(x<a\) and \(f''(x)>0\) when \(x>a\).
Note: Inflection point does not occur where \(f''(x) = 0\). The inflection point occurs where the sign of the second derivative change.
Question: Find the maximum and minimum value of the function \(x^5 - 5x^4 + 5 x^3 - 1\). Also, find the inflection point if any.
Solution: Let,
\begin{align*}
& f(x) = x^5 - 5x^4 + 5x^3 - 1 \\
& f'(x) = 5x^4 - 20x^3 + 15x^2
\end{align*}
For maximum or minimum \(f'(x) = 0\)
\begin{align*}
& 5x^4 - 20x^3 + 15x^2 = 0\\
or, ~ & 5x^2(x^2 - 4x + 3 ) = 0 \\
or, ~ & 5x^2(x - 1) (x - 3) = 0 \\
or, ~ & x = 0, 1,3
\end{align*}
Now, \(f''(x) = 20x^3 - 60x^2 + 30x\)
For x = 1
\(f''(x)= 20 - 60 +30 = -10 <0\)
then the function \(f(x)\) is maximum at x = 1 and the maximum value is, \begin{align*}
f(x) = & 1^5 - 5 \times 1^4 + 5 \times 1^3 - 1\\
= & 0
\end{align*}
For x = 3
\(f''(x) = 20 \times 3^3 - 60 \times 3^3 + 30 \times 3 = 90 >0\)
then the function \(f(x)\) is minimum at x = 3 and the minimum value is,
\begin{align*}
f(x) = & x^5 - 5\times 3^4 + 5 \times 3^3 - 1 \\
= & - 28
\end{align*}
For x = 0
\begin{align*}
f'' (x) & = 20 \times 0 - 60 \times 0 + 30 \times 0 \\
& = 0 \\
f'' (x) & = 20 x^3 - 60 x pe 2 + 30x \\
f''' (x) & = 60x^3 - 120x + 30 \\
\textrm{Now for}~~~ & x = 0, f'''(x) = 30 \ne 0
\end{align*}
the function has neither maximum nor minimum at x = 0
At x = 0 is an inflection point
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