Master Advanced Calculus Concepts with Step-by-Step Guidance
Calculus questions with fully worked solutions are presented below. Applications of the first and second derivatives include determining intervals of increase and decrease, local maxima and minima, concavity, and points of inflection.
For the function \[ f(x) = (x + 3)(x - 2)^3 \]
find:
a) First, we find the first derivative of the function using the product rule and the chain rule:
\[ f'(x) = (1)(x - 2)^3 + 3(x + 3)(x - 2)^2 \]
Factor out the common term \( (x - 2)^2 \):
\[ f'(x) = (x - 2)^2 [ (x - 2) + 3(x + 3) ] \]
\[ f'(x) = (x - 2)^2 (x - 2 + 3x + 9) \]
\[ f'(x) = (x - 2)^2(4x + 7) \]
Critical points are obtained by setting \( f'(x) = 0 \):
\[ x = 2 \quad \text{and} \quad x = -\dfrac{7}{4} \]
The sign table of \( f'(x) \) is shown below to determine intervals of increase and decrease:
| \(x\) | \((-\infty,-\frac{7}{4})\) | \(-\frac{7}{4}\) | \((-\frac{7}{4},2)\) | \(2\) | \((2,+\infty)\) |
|---|---|---|---|---|---|
| \((x-2)^2\) | \(+\) | \(+\) | \(+\) | \(0\) | \(+\) |
| \(4x+7\) | \(-\) | \(0\) | \(+\) | \(+\) | \(+\) |
| \(f'(x)\) | \(-\) | \(0\) | \(+\) | \(0\) | \(+\) |
| \(f(x)\) | \(\searrow\) (Decreasing) | Local Min | \(\nearrow\) (Increasing) | No Extremum | \(\nearrow\) (Increasing) |
From the sign table, \( f \) is decreasing on \( (-\infty, -\frac{7}{4}) \) and increasing on \( (-\frac{7}{4}, +\infty) \).
b) Since \( f'(x) \) changes sign from negative to positive at \( x = -\frac{7}{4} \), the function has a local minimum at:
\[ \mathbf{x = -\dfrac{7}{4}} \]
Although \( f'(2) = 0 \), there is no local extremum at \( x = 2 \) because the sign of \( f'(x) \) does not change (it remains positive on both sides).
c) To find concavity and inflection points, we calculate the second derivative \( f''(x) \) using the product rule on \( f'(x) = (x - 2)^2(4x + 7) \):
\[ f''(x) = 2(x - 2)(4x + 7) + 4(x - 2)^2 \]
Factor out the common term \( 2(x - 2) \):
\[ f''(x) = 2(x - 2) [ (4x + 7) + 2(x - 2) ] \]
\[ f''(x) = 2(x - 2) (4x + 7 + 2x - 4) \]
\[ f''(x) = 2(x - 2)(6x + 3) = 6(x - 2)(2x + 1) \]
Setting \( f''(x) = 0 \), the possible inflection points occur at \( x = 2 \) and \( x = -\frac{1}{2} \).
The sign table of \( f''(x) \) is shown below:
| \(x\) | \((-\infty,-\frac{1}{2})\) | \(-\frac{1}{2}\) | \((-\frac{1}{2},2)\) | \(2\) | \((2,+\infty)\) |
|---|---|---|---|---|---|
| \((x-2)\) | \(-\) | \(-\) | \(-\) | \(0\) | \(+\) |
| \((2x+1)\) | \(-\) | \(0\) | \(+\) | \(+\) | \(+\) |
| \(f''(x)\) | \(+\) | \(0\) | \(-\) | \(0\) | \(+\) |
| Concavity | Concave Up | Inflection | Concave Down | Inflection | Concave Up |
The graph of \( f \) is:
Given the function \[ f(x) = e^{x^2 - x} \]
find:
a) Find the first derivative using the chain rule. The derivative of \( e^{u} \) is \( e^{u} \cdot u' \):
\[ f'(x) = (2x - 1)e^{x^2 - x} \]
To find the critical points, set \( f'(x) = 0 \). Since exponential functions are strictly positive for all real numbers (\( e^{x^2 - x} > 0 \)), the only zero comes from the linear expression:
\[ 2x - 1 = 0 \implies x = \dfrac{1}{2} \]
The sign table of \( f'(x) \) is shown below:
| \(x\) | \((-\infty,\frac{1}{2})\) | \(\frac{1}{2}\) | \((\frac{1}{2},+\infty)\) |
|---|---|---|---|
| \(e^{x^2-x}\) | \(+\) | \(+\) | \(+\) |
| \(2x-1\) | \(-\) | \(0\) | \(+\) |
| \(f'(x)\) | \(-\) | \(0\) | \(+\) |
| \(f(x)\) | Decreasing | Local Min | Increasing |
The function is decreasing on \( (-\infty, \frac{1}{2}) \) and increasing on \( (\frac{1}{2}, +\infty) \).
b) Since \( f'(x) \) changes sign from negative to positive at \( x = \frac{1}{2} \), the function has a local minimum at:
\[ \mathbf{x = \dfrac{1}{2}} \]
c) To find concavity, calculate the second derivative \( f''(x) \) using the product rule and the chain rule:
\[ f''(x) = (2)e^{x^2 - x} + (2x - 1)(2x - 1)e^{x^2 - x} \]
Factor out the common term \( e^{x^2 - x} \):
\[ f''(x) = \big[2 + (2x - 1)^2\big] e^{x^2 - x} \]
Since the square of any real number is non-negative (\( (2x - 1)^2 \geq 0 \)), the expression inside the brackets satisfies \( 2 + (2x - 1)^2 \geq 2 \), making it strictly positive. As established, \( e^{x^2 - x} > 0 \).
Therefore, \( f''(x) > 0 \) for all real numbers \( x \). The graph is concave up on \( (-\infty, +\infty) \) and has no inflection points.