Point of Inflection - Point of Concavity Change

Solution

We follow the 3 steps written above:

• Step 1: find \(f''(x)\). Differentiating with respect to \(x\): \[f'(x)=3x^2-4x - 8\] Differentiating again: \[f''(x) = 6x-4\]

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• Step 2: Solve \(f''(x)=0\): \[\begin{aligned} & 6x - 4 = 0 \\ & 6x = 4 \\ & x = \frac{4}{6} \\ & x = \frac{2}{3} \end{aligned}\]

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• Step 3: check whether \(f''(x)\) changes sign.
  • for \(x< \frac{2}{3}\), \(f''(x)<0\)
  • for \(x> \frac{2}{3}\), \(f''(x)>0\)
  Since \(f''\begin{pmatrix}\frac{2}{3}\end{pmatrix}=0\) and \(f''(x)\) changes sign on either side of \(x=\frac{2}{3}\) we can conclude that there is a point of inflection when \(x=\frac{2}{3}\).

Looking at this function's graph, we can see quite clearly that the curve has a point of inflection with coordinates \(\begin{pmatrix} \frac{2}{3},-\frac{160}{27}\end{pmatrix}\); indeed there is a concavity change on either side of this point:

• concave down for \(x<\frac{2}{3}\) (to the left of the point)
• concave up for \(x>\frac{2}{3}\) (to the right of the point)

Solution

We follow the 3 steps written further up:

• Step 1: find \(f''(x)\). Differentiating with respect to \(x\): \[f'(x)=-2x^3 + 12x^2+30x + 2\] Differentiating again: \[f''(x)=-6x^2+24x+30\]

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• Step 2: Solve \(f''(x)=0\). \[\begin{aligned} & -6x^2+24x+30 = 0 \\ & -6\begin{pmatrix} x^2 - 4x - 5 \end{pmatrix} = 0 \\ & - 6 \begin{pmatrix}x + 1 \end{pmatrix} \begin{pmatrix}x-5 \end{pmatrix} = 0 \end{aligned}\] We can see that \(f''(x) = 0\) has 2 solutions: \[x = -1 \quad \text{and} \quad x = 5\]

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• Step 3: Check that \(f''(x)\) changes sign on either side of these values of \(x\).
  To study the sign of \(f''(x)\), on either side of \(x=-1\) and \(x=5\), we can construct a sign table (as shown here). Using \(f''(x)=- 6 \begin{pmatrix}x + 1 \end{pmatrix} \begin{pmatrix}x-5 \end{pmatrix}\) we study the sign of each of the factors, \(-6\), \(x+1\) and \(x-5\), to then study the overall sign of \(f''(x)\).

  
  On the last row of this sign table, we can see that (as we go from left to right) \(f''(x)\) changes from negative to positive across \(x=-1\) and from positive to negative across \(x=5\). This confirms that the points with \(x\) coordinates \(x=-1\) and \(x=5\) are indeed points of inflection.

These results are further confirmed by plotting the curve \(y=-\frac{x^4}{2} + 4x^3 + 15x^2 + 2x + 1\). We can see the points of inflection, found above, clearly marking a change in concavity on this curve:

• for \(x < -1 \): the curve is concave-down and \(f''(x)<0\)
• for \(- 1 < x < 5\): the curve is concave-up and \(f''(x)>0\)
• for \(x > 5\): the curve is concave-down and \(f''(x)<0\)

Solution

Following our 3 step method:

• Step 1: find \(f''(x)\) by successive differentiation \[f'(x) = cos(x)\] differentiating once more: \[f''(x) = -sin(x)\]

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• Step 2: solve \(f''(x)=0\). Since \f''(x)=-sin(x)\), this results in solving: \[-sin(x)=0, \quad -\frac{3 \pi}{2} \leq x \leq \frac{5 \pi}{2} \] which leads to the 4 solutions: \[x = -\pi, \ 0, \ \pi, \ 2\pi\]

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• Step 3: check whether \(f''(x)\) changes sign.
  To do this we can study the sign of \(f''(x)\), writing it as \(f''(x)=-1\times sin(x)\) in a sign table, as shown here.

  
  The bottom row of this table clearly shows that \(f''(x)\) changes sign on either side of each of the values \(x = -\pi, \ 0, \ \pi, \ 2\pi\). Each of these values of \(x\) therefore corresponds to a point of inflection.

Since \(sin\begin{pmatrix}-\pi \end{pmatrix} = sin\begin{pmatrix}0 \end{pmatrix} = sin\begin{pmatrix}\pi \end{pmatrix} = sin\begin{pmatrix}2\pi \end{pmatrix} = 0\)

\(\begin{pmatrix}-\pi,\ 1.2 \end{pmatrix}\), \(\begin{pmatrix}0,\ 1.2 \end{pmatrix}\), \(\begin{pmatrix}\pi,\ 1.2 \end{pmatrix}\) and \(\begin{pmatrix}2\pi,\ 1.2 \end{pmatrix}\).