Concavity and Inflection Points – Practice Questions

Below are calculus practice questions with detailed solutions on concavity[cite: 1] and inflection points of function graphs. Each solution is hidden inside a collapsible dropdown so you can practice independently before reviewing the steps.

Questions and Solutions

Question 1

Determine the concavity of the graph of the general quadratic function[cite: 1] defined by \[ f(x) = ax^2 + bx + c \]

View Solution
  • Compute the derivatives: \[ f'(x) = 2ax + b \] \[ f''(x) = 2a \]
  • The sign of \( f''(x) \) depends only on \( a \):
    • If \( a > 0 \), the graph is concave up.
    • If \( a < 0 \), the graph is concave down.

Question 2

Find all intervals where the function \[ f(x) = \sin x \] is concave up.

View Solution
  • \[ f'(x) = \cos x, \qquad f''(x) = -\sin x \]
  • Concavity up occurs when \[ f''(x) > 0 \quad \Longrightarrow \quad \sin x < 0 \]
  • Over one period \( [0, 2\pi] \), this happens on \( (\pi, 2\pi) \).
  • Therefore, \[ (\pi + 2\pi k,\; 2\pi + 2\pi k), \quad k \in \mathbb{Z} \]

Question 3

The graph of \( f'(x) \) is shown below for \( x \in [a,g] \). On which intervals is \( f \) decreasing and concave down? Also find all inflection points.

Graph of the first derivative

View Solution
  • \( f \) is decreasing where \( f'(x) < 0 \).
  • \( f \) is concave down where \( f'(x) \) is decreasing.

Sign table of derivatives

  • Decreasing and concave down on: \[ (0,d) \quad \text{and} \quad (e,g) \]
  • Inflection points: \[ (c,f(c)),\ (d,f(d)),\ (e,f(e)),\ (f,f(f)) \]

Question 4

Find all inflection points of \[ f(x) = 4x^4 - x^3 + 2 \]

View Solution
  • \[ f'(x) = 16x^3 - 3x^2 \] \[ f''(x) = 48x^2 - 6x = 6x(8x - 1) \]
  • Sign changes occur at: \[ x = 0, \quad x = \frac{1}{8} \]
  • Inflection points: \[ (0,2), \quad \left(\frac{1}{8}, \frac{2047}{1024}\right) \]

Concavity sign chart

Question 5

Determine the inflection point of \[ f(x) = -x^3 + 3x^2 + 1 \] Then find the inflection point of \[ g(x) = -(x-2)^3 + 3(x-2)^2 + 1 \]

View Solution
  • \[ f'(x) = -3x^2 + 6x, \quad f''(x) = -6x + 6 \]
  • \[ f''(x) = 0 \Rightarrow x = 1 \]
  • Inflection point of \(f\): \[ (1,3) \]
  • Since \( g(x) = f(x-2) \), the graph shifts right 2 units.
  • Inflection point of \(g\): \[ (3,3) \]

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