The Second Derivative Test

1. Using the power rule for differentiation we find: \[\begin{aligned} f'(x) & = 3x^{3-1} + 3\times 2x^{2-1}-9x^{1-1} + 0 \\ & = 3x^2+3\times 2x^1-9x^0 \\ f'(x) & = 3x^2 + 6x - 9 \end{aligned}\]

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2. The \(x\) coordinates of this function's stationary points are found by solving \(f'(x) = 0\). So we solve: \[\begin{aligned} & f'(x) = 0 \\ & 3x^2 + 6x - 9 = 0 \\ & 3\begin{pmatrix}x^2 + 2x - 3 \end{pmatrix} = 0 \\ & 3 \begin{pmatrix} x+3 \end{pmatrix}\begin{pmatrix} x - 1 \end{pmatrix} = 0 \end{aligned}\] This quadratic equation has two solutions \(x = -3\) and \(x = 1\). To find the \(y\) coordinates of the stationary points we calculate \(f(-3)\) and \(f(1)\), leading to:
   \[\begin{aligned} f(-3) & = \begin{pmatrix}-3\end{pmatrix}^3 + 3\begin{pmatrix}-3\end{pmatrix}^2 - 9\begin{pmatrix}-3\end{pmatrix} + 24 \\ & = -27 + 3\times 9 - (-27)+24 \\ & = -27 + 27 +27 + 24 \\ f(-3) & = 51 \end{aligned}\]

   \[\begin{aligned} f(1) & = 1^3 + 3\times 1^2 - 9\times 1 + 24 \\ & = 1 +3\times 1 - 9 + 24 \\ & = 1 + 3 - 9 + 24 \\ f(1) & = 19 \end{aligned}\]
   The function has two stationary points: \[\begin{pmatrix}-3,51\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}1,19 \end{pmatrix}\]

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3. We differentiate \(f'(x) = 3x^2 + 6x - 9\) to find \(f''(x)\): \[\begin{aligned} f''(x) & = 3\times 2x^{2-1} + 6 \\ & = 6x^1 + 6 \\ f''(x) & = 6x + 6 \end{aligned}\]

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4. We calculate, and study the sign of, the second derivative at each stationary point:
   • at \(\begin{pmatrix}-3,-51\end{pmatrix}\): \[\begin{aligned} f''(-3) & = 6\times (-3) + 6 \\ & = -18 + 6 \\ f''(-3) & = -12 \end{aligned}\] Since \(f''(-3)<0\) the stationary point \(\begin{pmatrix}-3,51\end{pmatrix}\) is a maximum.
   • at \(\begin{pmatrix}1,19 \end{pmatrix}\): \[\begin{aligned} f''(1) & = 6 \times 1 + 6 \\ & = 6 + 6 \\ f''(1) & = 12 \end{aligned}\] Since \(f''(1)>0\) the stationary point \(\begin{pmatrix}1,19\end{pmatrix}\) is a minimum.

The function we're working with is \(y = -2x^3 + 9x^2 + 60x + 6\)

1. Using the power rule for differeentiation we find: \[\begin{aligned} \frac{dy}{dx} & = -2 \times 3x^{3-1} + 9\times 2x^{2-1} + 60 + 0 \\ & = -6x^2 + 18x^1 + 60 \\ \frac{dy}{dx} & = -6x^2 + 18x + 60 \end{aligned}\]
2. To find the \(x\) coordinates of any stationary points, we solve \(\frac{dy}{dx} = 0\): \[\begin{aligned} & \frac{dy}{dx} = 0 \\ & -6x^2 + 18x + 60 = 0 \\ & -6 \begin{pmatrix} x^2 - 3x - 10 \end{pmatrix} = 0 \\ & -6\begin{pmatrix}x+2\end{pmatrix}\begin{pmatrix}x-5 \end{pmatrix} = 0 \end{aligned}\] This leads to two solutions and therefore two stationary points whose \(x\) coordinates are \(x = -2\) and \(x = 5\). To find their \(y\) coordinates we use the function we were given in the question:
   When \(x = -2\): \[\begin{aligned} y & = -2(-2)^3 + 9(-2)^2 + 60(-2)+6 \\ & = -2(-8) + 9\times 4 + (-120)+6 \\ & = 16 + 36 -120 + 6 \\ y & = -62 \end{aligned}\]

   When \(x = 5\): \[\begin{aligned} y & = -2\times 5^3 + 9 \times 5^2 + 60 \times 5 + 6 \\ & = -2\times 125 + 9 \times 25 + 300 +6 \\ & = -250 + 225 + 300 + 6 \\ y & = 281 \end{aligned}\]
   So this function two stationary points' coordinates are: \[\begin{pmatrix}-2,-62\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}5,281\end{pmatrix}\]
3. Starting from \(\frac{dy}{dx} = -6x^2 + 18x + 60\) and using the power rule for differentiation, we find: \[\begin{aligned} \frac{d^2y}{dx^2} & = -6\times 2x^{2-1} + 18 + 0 \\ & = -12x^1 + 18 \\ \frac{d^2y}{dx^2}& = -12x + 18 \end{aligned}\]
4. We evaluate the second derivative at each of the stationary points:
   • at \(\begin{pmatrix}-2,-62\end{pmatrix}\): \[\begin{aligned} \frac{d^2y}{dx^2} = & = -12 \times (-2) + 18 \\ & = 24 + 18 \\ \frac{d^2y}{dx^2} & = 42 \end{aligned}\] Since \(\frac{d^2y}{dx^2}>0\) the stationary point \(\begin{pmatrix}-2,-62\end{pmatrix}\) is a minimum
   • at \(\begin{pmatrix}5,281\end{pmatrix}\): \[\begin{aligned} \frac{d^2y}{dx^2} = & = -12 \times 5 + 18 \\ & = -60 + 18 \\ \frac{d^2y}{dx^2} & = -42 \end{aligned}\] Since \(\frac{d^2y}{dx^2} <0 \) the stationary point \(\begin{pmatrix}5,281\end{pmatrix}\) is a maximum.

Given the function \(f(x) = x^2.e^x\) we find:

1. To find the derivative \(f'(x)\) we use the product rule. Let: \[u(x) = x^2 \quad \text{and} \quad v(x) = e^x\] then: \[u'(x) = 2x \quad \text{and} \quad v'(x) = e^x\] Using the product rule: \[\begin{aligned} f'(x) &= u'(x)v(x) + u(x)v'(x) \\ & = 2x.e^x + x^2.e^x \\ f'(x) & = e^x\begin{pmatrix}2x + x^2 \end{pmatrix} \end{aligned}\]

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2. We find the \(x\) coordinates of any stationary points by solving \(f'(x)=0\), that's: \[\begin{aligned} f'(x) & = 0 \\ e^x\begin{pmatrix}2x + x^2 \end{pmatrix} & = 0 \\ e^x.x.\begin{pmatrix} 2 + x \end{pmatrix} & = 0 \end{aligned}\] Since \(e^x \neq 0\), this equation has two solutions \(x = 0\) and \(x = -2\). The function therefore has two stationary points with \(x\) coordinates \(x=-2\) and \(x=0\).
   We calculate the \(y\) coordinates of the stationary point using the function \(f(x) = x^2.e^x\):
   when \(x = -2\): \[\begin{aligned} f(-2) &= (-2)^2.e^{-2} \\ & = 4 \times \frac{1}{e^2} \\ f(-2) &= \frac{4}{e^2} \end{aligned}\]

   when \(x = 0\): \[\begin{aligned} f(0) &= 0^2.e^0 \\ & = 0 \times 1 \\ f(0) &= 0 \end{aligned}\]
   This function's two stationary points are: \[\begin{pmatrix}-2,\frac{4}{e^2}\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}0,0\end{pmatrix}\]

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3. Starting from \(f'(x) = e^x\begin{pmatrix}2x + x^2 \end{pmatrix}\) and using the product rule we find \(f''(x)\). Let: \[u(x)=e^x\quad \text{and} \quad v(x)=2x+x^2\] then: \[u'(x)=e^x\quad \text{and} \quad v'(x)=2+2x\] and \[\begin{aligned} f''(x) & = u'(x)v(x) + u(x)v'(x) \\ & = e^x\begin{pmatrix}2x+x^2\end{pmatrix} + e^x\begin{pmatrix}2+2x \end{pmatrix} \\ f''(x) & = e^x\begin{pmatrix}x^2 + 4x + 2 \end{pmatrix} \end{aligned}\]

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4. We use the second derivative test at each of the stationary points to determine their nature:
   • at \(\begin{pmatrix}-2,\frac{4}{e^2}\end{pmatrix}\): \[\begin{aligned} f''(-2) & = e^{-2}\begin{pmatrix}(-2)^2 + 4\times (-2) + 2 \end{pmatrix} \\ & = e^{-2}\begin{pmatrix}4 - 8 +2 \end{pmatrix} \\ & = -2.e^{-2} \\ f''(-2) & = - \frac{2}{e^2} \end{aligned}\] \(f''(-2)<0\) the stationary point is a maximum.
   • at \(\begin{pmatrix}0,0\end{pmatrix}\): \[\begin{aligned} f''(0) & = e^0\begin{pmatrix}0^2 + 4\times 0 + 2 \end{pmatrix} \\ & = 1 \times 2 \\ f''(0) & = 2 \end{aligned}\] \(f''(0)>0\) the stationary point is a minimum.

Working with the function \(f(x) = 2x^3+18x^2 + 54x + 49\):

1. To find the derivative \(f'(x)\) we use the power rule: \[\begin{aligned} f'(x) & = 2\times 3 x^{3-1} + 18\times 2 x^{2-1} + 54 + 0 \\ & = 6x^2 + 36x^1 + 54 \\ f'(x) & = 6x^2 + 36x + 54 \end{aligned}\]

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2. We find the \(x\) coordinates of any stationary points by solving \(f'(x)=0\), that's: \[\begin{aligned} f'(x) & = 0 \\ 6x^2 + 36x + 54 = 0 \\ 6\begin{pmatrix}x^2 + 6x + 9 \end{pmatrix} = 0 \\ 6\begin{pmatrix} x + 3 \end{pmatrix}^2 = 0 \end{aligned}\] Since this equation has one single solution \(x = -3\) . The function therefore has one stationary point with \(x\) coordinates \(x=-3\).
   We calculate the \(y\) coordinates of the stationary point using the function \(f(x) = 2x^3+18x^2 + 54x + 49\): \[\begin{aligned} f(-3) & = 2(-3)^2 + 18(-3)^2+54(-3)+49 \\ & = 2(-27) + 18 \times 9 + (-162) + 49 \\ & = -54 + 162 -162 + 49 \\ f(-3) & = -5 \end{aligned}\] This function's stationary point has coordinates: \[\begin{pmatrix}-3,-5\end{pmatrix}\]

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3. Starting from \(f'(x) = 6x^2 + 36x + 54\) and using the power rule we find \(f''(x)\): \[\begin{aligned} f''(x) & = 6\times 2x^{2-1} + 36 \\ & = 12x^1 + 36 \\ f''(x) & = 12x + 36 \end{aligned}\]

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4. We try using the second derivative test at the stationary point \(\begin{pmatrix}-3,-5\end{pmatrix}\): \[\begin{aligned} f''(-3) & = 12 \times (-3) + 36 \\ & = -36 + 36 \\ f''(-3) & = 0 \end{aligned}\] Since \(f''(-3)=0\) the second derivative test is inconclusive and we have to study the sign of the first derivative \(f'(x)\) on either side of the stationary point. To to this we write \(f'(x)\) in its factored form and study its sign with a sign table: \[f'(x) = 6\begin{pmatrix}x+3\end{pmatrix}^2\]
   Each of the factors, \(6\) and \(\begin{pmatrix}x+3\end{pmatrix}^2\), is a row in the table. We then use a row for \(f'(x)\) as well as a row to illustrate whether \(f(x)\) increases or decreases:

   • \(6\) is always positive
   • \(\begin{pmatrix}x+3\end{pmatrix}^2\) equals to \(0\) when \(x = -3\) and is positive for all other values of \(x\).
   • since \(f'(x)\) is equal to the product of the factors, its sign is equal to the product of the signs in each of the rows.
   • \(f(x)\) increases when \(f'(x)>0\), decreases if \(f'(x)<0\), and reaches a stationary point if \(f'(x)=0\).

   
   Since \(f'(x)>0\) on either side of the stationary point \(\begin{pmatrix}-3,-5\end{pmatrix}\) the stationary point is a increasing horizontal point of inflection.