Session 09-05: Tasks

Final Recap - Calculus & Curve Sketching

Exam-Style Problem: Graphical Differentiation / Integration / System of Linear Equations

Consider the following third degree polynomial \(g(x)\):

Figure 1: Graph of the third degree polynomial g(x)

Part a)

Graph the derivative \(g'(x)\) of the function depicted above directly into the figure.

Key observations for sketching \(g'(x)\):

  • \(g'(x) = 0\) at \(x = 0\) (local max of \(g\); double root touch) and \(x = 4\) (local min of \(g\))
  • \(g'(x) > 0\) for \(x < 0\) and \(x > 4\) (where \(g\) is increasing)
  • \(g'(x) < 0\) for \(0 < x < 4\) (where \(g\) is decreasing)
  • \(g'(x)\) is a quadratic (upward-opening parabola) since \(g\) has a positive leading coefficient
Figure 2: Graph of g(x) with its derivative g’(x)

Part b)

Graph the antiderivative \(G(x)\) of the function depicted above directly into the figure and interpret its relation to \(g(x)\).

Since \(G'(x) = g(x)\), the antiderivative \(G\) increases where \(g > 0\) and decreases where \(g < 0\).

Sign analysis of \(g(x) = \frac{1}{4}x^2(x-6)\):

  • For \(x < 0\): \(x^2 > 0\) and \((x-6) < 0\), so \(g(x) < 0\) \(\Rightarrow\) \(G\) decreasing
  • For \(0 < x < 6\): \(x^2 > 0\) and \((x-6) < 0\), so \(g(x) < 0\) \(\Rightarrow\) \(G\) decreasing
  • For \(x > 6\): \(x^2 > 0\) and \((x-6) > 0\), so \(g(x) > 0\) \(\Rightarrow\) \(G\) increasing

Special points:

  • \(g(0) = 0\) but \(g\) does not change sign (double root) \(\Rightarrow\) \(G\) has an inflection point at \(x = 0\)
  • \(g(6) = 0\) and \(g\) changes sign \(\Rightarrow\) \(G\) has a local minimum at \(x = 6\)

Choosing \(G(0) = 0\):

\[G(x) = \frac{x^4}{16} - \frac{x^3}{2}\]

Figure 3: Graph of g(x) with its antiderivative G(x) (scaled by 1/4 for visibility)

Interpretation: \(G(x)\) represents the accumulated (signed) area under \(g(x)\). Where \(g(x) > 0\), \(G\) increases; where \(g(x) < 0\), \(G\) decreases. The zeros of \(g\) correspond to critical points or inflection points of \(G\), depending on whether \(g\) changes sign.

Part c)

Determine the functional equation \(g(x) = ax^3 + bx^2 + cx + d\) by determining the values of \(a\), \(b\), \(c\), and \(d\).

From the graph we read the following information:

  • \(g(0) = 0\) (passes through origin)
  • \(g'(0) = 0\) (local maximum at the origin; the graph touches the \(x\)-axis)
  • \(g'(4) = 0\) (local minimum at \(x = 4\))
  • \(g(4) = -8\) (value at the local minimum)

From \(g(0) = 0\): \(d = 0\)

From \(g'(0) = 0\): \(g'(x) = 3ax^2 + 2bx + c\), so \(g'(0) = c = 0\)

With \(c = 0\) and \(d = 0\), we have \(g(x) = ax^3 + bx^2\) and \(g'(x) = 3ax^2 + 2bx\).

Setting up the remaining equations:

\[g'(4) = 48a + 8b = 0 \quad \Rightarrow \quad 6a + b = 0 \quad \Rightarrow \quad b = -6a \quad (I)\]

\[g(4) = 64a + 16b = -8 \quad (II)\]

Substituting (I) into (II): \(64a + 16(-6a) = -8\)

\[64a - 96a = -8 \quad \Rightarrow \quad -32a = -8 \quad \Rightarrow \quad a = \frac{1}{4}\]

Back-substituting: \(b = -6 \cdot \frac{1}{4} = -\frac{3}{2}\)

\[\boxed{g(x) = \frac{1}{4}x^3 - \frac{3}{2}x^2}\]

Verification: \(g(x) = \frac{1}{4}x^2(x - 6)\)

  • \(g(0) = 0\) \(\checkmark\)
  • \(g(6) = \frac{1}{4}(36)(0) = 0\) \(\checkmark\)
  • \(g(4) = \frac{1}{4}(16)(-2) = -8\) \(\checkmark\)

Part d)

Determine the functional equation of the tangent \(t(x)\) at \(x = -2\).

(Hint: Continue your work using the function \(g(x) = \frac{1}{4}x^3 - \frac{3}{2}x^2\).)

Using the function \(g(x) = \frac{1}{4}x^3 - \frac{3}{2}x^2\):

Function value at \(x = -2\):

\[g(-2) = \frac{-8}{4} - \frac{3 \cdot 4}{2} = -2 - 6 = -8\]

Slope at \(x = -2\):

\[g'(x) = \frac{3x^2}{4} - 3x\]

\[g'(-2) = \frac{3 \cdot 4}{4} - 3 \cdot (-2) = 3 + 6 = 9\]

Tangent equation:

\[t(x) = g'(-2)(x - (-2)) + g(-2) = 9(x + 2) - 8 = 9x + 18 - 8\]

\[\boxed{t(x) = 9x + 10}\]

Part e)

Compute the angle of intersection \(\alpha\) between the tangent and the \(x\)-axis.

The slope of the tangent is \(m = 9\).

\[\tan(\alpha) = m = 9\]

\[\alpha = \arctan(9) \approx 83.66°\]

\[\boxed{\alpha \approx 83.7°}\]