Calculus 1 Practice Questions - Set A

Practice Questions and Detailed Solutions

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1. Question 1
   Find the domain of the function: \[ f(x) = \frac{\sqrt{x - 1}}{\sqrt{4 - x^2}} \]
   Show Solution to Question 1

   For function \( f \) to take real values, the expression under the radical in the numerator must be non-negative, and the expression under the radical in the denominator must be positive. We solve the inequalities:

   \[ x - 1 \ge 0 \quad \text{and} \quad 4 - x^2 > 0 \]

   The solution sets for the first and second inequalities are respectively:

   \[ x \ge 1 \quad \text{and} \quad -2 < x < 2 \]

   Both conditions must be satisfied simultaneously. Therefore, the domain is the intersection of these two sets:

   \[ 1 \le x < 2 \]
2. Question 2
   Find the range of the function: \[ f(x) = \frac{x - 1}{2 - 3x} \]
   Show Solution to Question 2

   According to the properties of inverse functions, the range of \( f \) is the domain of its inverse.

   First, prove that \( f \) is a one-to-one function by assuming \( f(a) = f(b) \) and showing \( a = b \):

   \[ \frac{a - 1}{2 - 3a} = \frac{b - 1}{2 - 3b} \]

   Cross-multiplying and simplifying gives \( a = b \), proving invertibility. To find the inverse, start with:

   \[ y = \frac{x - 1}{2 - 3x} \implies 2y - 3xy = x - 1 \implies x = \frac{2y + 1}{3y + 1} \]

   Interchanging \( x \) and \( y \) gives the inverse function:

   \[ f^{-1}(x) = \frac{2x + 1}{3x + 1} \]

   The domain of \( f^{-1} \) excludes \( x = -\dfrac{1}{3} \). Thus, the range of \( f \) in interval notation is:

   \[ \left(-\infty, -\frac{1}{3}\right) \cup \left(-\frac{1}{3}, +\infty\right) \]
3. Question 3
   Find the inverse of the function: \[ f(x) = \ln(2x - 3) + 2 \]
   Show Solution to Question 3

   Write the function as an equation and solve for \( x \):

   \[ y = \ln(2x - 3) + 2 \] \[ y - 2 = \ln(2x - 3) \] \[ e^{y - 2} = 2x - 3 \implies 2x = e^{y - 2} + 3 \implies x = \frac{1}{2}\left(e^{y - 2} + 3\right) \]

   Interchanging \( x \) and \( y \) gives the inverse function:

   \[ f^{-1}(x) = \frac{1}{2}\left(e^{x - 2} + 3\right) \]
4. Question 4
   Evaluate each of the following limits:
   • \( \displaystyle \lim_{x \to 16} \frac{-\frac{1}{\sqrt{x}} + \frac{1}{4}}{x - 16} \)
   • \( \displaystyle \lim_{x \to +\infty} \frac{-x^3 + 2x - 1}{x^4 - 3x^3 + 9} \)
   • \( \displaystyle \lim_{x \to +\infty} x \sin\left(\frac{3}{x}\right) \)
   • \( \displaystyle \lim_{x \to 0} \frac{\sin(x) + x}{2x^2 + x} \)
   • \( \displaystyle \lim_{x \to +\infty} \frac{\sin(x) + 1}{x} \)
   Show Solution to Question 4

   a) The limit is of the indeterminate form \( \dfrac{0}{0} \). Multiply numerator and denominator by the conjugate of the numerator:

   \[ \lim_{x \to 16} \frac{-\frac{1}{\sqrt{x}} + \frac{1}{4}}{x - 16} = \lim_{x \to 16} \frac{\left(-\frac{1}{\sqrt{x}} + \frac{1}{4}\right)\left(-\frac{1}{\sqrt{x}} - \frac{1}{4}\right)}{(x - 16)\left(-\frac{1}{\sqrt{x}} - \frac{1}{4}\right)} = \lim_{x \to 16} \frac{\frac{1}{x} - \frac{1}{16}}{(x - 16)\left(-\frac{1}{\sqrt{x}} - \frac{1}{4}\right)} \] \[ = \lim_{x \to 16} \frac{\frac{16 - x}{16x}}{(x - 16)\left(-\frac{1}{\sqrt{x}} - \frac{1}{4}\right)} = \lim_{x \to 16} \frac{-1}{16x\left(-\frac{1}{\sqrt{x}} - \frac{1}{4}\right)} = \frac{-1}{16(16)\left(-\frac{1}{4} - \frac{1}{4}\right)} = \frac{1}{128} \]

   b) The limit is of the form \( \dfrac{\infty}{\infty} \). Divide all terms by the highest power \( x^4 \):

   \[ \lim_{x \to +\infty} \frac{-x^3 + 2x - 1}{x^4 - 3x^3 + 9} = \lim_{x \to +\infty} \frac{-\frac{1}{x} + \frac{2}{x^3} - \frac{1}{x^4}}{1 - \frac{3}{x} + \frac{9}{x^4}} = \frac{0}{1} = 0 \]

   c) Indeterminate form \( \infty \cdot 0 \). Let \( t = \dfrac{3}{x} \):

   \[ \lim_{x \to +\infty} x \sin\left(\frac{3}{x}\right) = \lim_{t \to 0} 3 \frac{\sin(t)}{t} = 3 \times 1 = 3 \]

   d) Indeterminate form \( \dfrac{0}{0} \). Apply L'Hôpital's Rule:

   \[ \lim_{x \to 0} \frac{\sin(x) + x}{2x^2 + x} = \lim_{x \to 0} \frac{\cos(x) + 1}{4x + 1} = \frac{\cos(0) + 1}{0 + 1} = 2 \]

   e) Since \( -1 \le \sin(x) \le 1 \), adding 1 gives \( 0 \le \sin(x) + 1 \le 2 \). Dividing by \( x > 0 \):

   \[ 0 \le \frac{\sin(x) + 1}{x} \le \frac{2}{x} \]

   By the Squeeze Theorem, since limits of the bounds are 0, the limit evaluates to \( 0 \).

5. Question 5
   a) Graph \( y = e^{x-1} \) and \( y = x \) in the same coordinate system, show that the two graphs are tangent at the point \( (1, 1) \), and verify that \( e^{x-1} \ge x \).
   b) Use the result from part a) to determine the concavity of the function \( f(x) = \dfrac{x^3}{6} - e^{x-1} \) and any inflection points.
   Show Solution to Question 5

   a) The derivative of \( y = e^{x-1} \) is \( y' = e^{x-1} \). At \( x = 1 \), slope \( m = e^{1-1} = 1 \). The tangent line at \( (1, 1) \) is \( y - 1 = 1(x - 1) \implies y = x \). Thus, the graphs are tangent at \( (1, 1) \) and \( e^{x-1} \ge x \).

   

   b) Given \( f(x) = \dfrac{x^3}{6} - e^{x-1} \), the derivatives are:

   \[ f'(x) = \frac{x^2}{2} - e^{x-1}, \quad f''(x) = x - e^{x-1} \]

   Since \( e^{x-1} \ge x \implies x - e^{x-1} \le 0 \), \( f''(x) \le 0 \) everywhere (with equality at \( x = 1 \)). Thus, \( f(x) \) is concave down on \( (-\infty, +\infty) \) and has no inflection points since \( f'' \) does not change sign.

6. Question 6
   Find the derivative of the following functions (do not simplify the final answer):
   • \( f(x) = e^{x-1} + \ln(3x - 1) + \sin(2x + 1) \)
   • \( g(x) = (2x - 1)^2(\tan(x) - 1) \)
   • \( h(x) = \dfrac{x - \cos(x)}{x^2 - 2x + 1} \)
   • \( m(x) = \sin\left(\sqrt{x^3 - \dfrac{1}{x} + 2}\right) \)
   • \( n(x) = 3^{2x+3} + \log_3(2x - 1) \)
   Show Solution to Question 6

   a) Sum rule: \( f'(x) = e^{x-1} + \dfrac{3}{3x - 1} + 2\cos(2x + 1) \)

   b) Product rule: \( g'(x) = 4(2x - 1)(\tan(x) - 1) + (2x - 1)^2\sec^2(x) \)

   c) Quotient rule: \( h'(x) = \dfrac{(1 + \sin(x))(x^2 - 2x + 1) - (x - \cos(x))(2x - 2)}{(x^2 - 2x + 1)^2} \)

   d) Chain rule: \( m'(x) = \cos\left(\sqrt{x^3 - \dfrac{1}{x} + 2}\right) \cdot \frac{1}{2}\left(3x^2 + \frac{1}{x^2}\right)\left(x^3 - \dfrac{1}{x} + 2\right)^{-1/2} \)

   e) Rewriting bases: \( n(x) = e^{(2x+3)\ln 3} + \dfrac{\ln(2x-1)}{\ln 3} \implies n'(x) = (2\ln 3)3^{2x+3} + \dfrac{2}{\ln 3(2x - 1)} \)

7. Question 7
   Find the equation of the tangent line to the curve given by \( \sin(y^2) = x^2 \) at the point \( (0, \sqrt{\pi}) \).
   Show Solution to Question 7

   Differentiating implicitly with respect to \( x \):

   \[ 2y \frac{dy}{dx}\cos(y^2) = 2x \implies \frac{dy}{dx} = \frac{x}{y\cos(y^2)} \]

   Evaluating at \( (0, \sqrt{\pi}) \):

   \[ m = \frac{0}{\sqrt{\pi}\cos(\pi)} = 0 \]

   The equation of the horizontal tangent line is:

   \[ y = \sqrt{\pi} \]
8. Question 8
   Find the constants \( a \) and \( b \) so that the function \( f \) is continuous on \( (-\infty, +\infty) \): \[ f(x) = \begin{cases} 2x - 1 & x \le 1 \\ ax^3 + b & 1 < x < 2 \\ x + 2b & x \ge 2 \end{cases} \]
   Show Solution to Question 8

   Continuity at \( x = 1 \):

   \[ \lim_{x \to 1^-} (2x - 1) = 1, \quad \lim_{x \to 1^+} (ax^3 + b) = a + b \implies a + b = 1 \quad \text{(Eq. 1)} \]

   Continuity at \( x = 2 \):

   \[ \lim_{x \to 2^-} (ax^3 + b) = 8a + b, \quad \lim_{x \to 2^+} (x + 2b) = 2 + 2b \implies 8a + b = 2 + 2b \quad \text{(Eq. 2)} \]

   Solving equations (1) and (2) simultaneously yields:

   \[ a = \frac{1}{3}, \quad b = \frac{2}{3} \]
9. Question 9
   Find the equation of the tangent line to the curve given by \( y = x + \sin(x) \) at \( x = 0 \).
   Show Solution to Question 9

   Derivative: \( y' = 1 + \cos(x) \). Slope at \( x = 0 \) is \( m = 1 + \cos(0) = 2 \).

   Point of tangency: \( P(0, 0 + \sin(0)) = (0, 0) \).

   Tangent line equation: \( y - 0 = 2(x - 0) \implies y = 2x \).

10. Question 10
    Use the definition of the derivative as a limit to find \( f'(x) \) where \( f(x) = \sqrt{x + 2} \).
    Show Solution to Question 10

    Using the limit definition of the derivative:

    \[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h + 2} - \sqrt{x + 2}}{h} \]

    Multiply by the conjugate:

    \[ f'(x) = \lim_{h \to 0} \frac{(x + h + 2) - (x + 2)}{h\left(\sqrt{x + h + 2} + \sqrt{x + 2}\right)} = \lim_{h \to 0} \frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}} = \frac{1}{2\sqrt{x + 2}} \]
11. Question 11
    Determine on what interval(s) the function \( f(x) = e^x(x^2 - 5x + 8) + \dfrac{x^4}{12} - \dfrac{x^3}{6} \) is concave up and concave down, and locate any inflection points.
    Show Solution to Question 11

    First derivative: \( f'(x) = e^x(x^2 - 3x + 3) + \dfrac{x^3}{3} - \dfrac{x^2}{2} \)

    Second derivative: \( f''(x) = e^x(x^2 - x) + x^2 - x = x(x - 1)e^x \)

    Zeros of \( f'' \) are \( x = 0 \) and \( x = 1 \). Testing intervals:

    • \( (-\infty, 0) \): \( f''(-1) > 0 \) (concave up)
    • \( (0, 1) \): \( f''(1/2) < 0 \) (concave down)
    • \( (1, +\infty) \): \( f''(2) > 0 \) (concave up)

    Points of inflection occur at \( x = 0 \) and \( x = 1 \).

12. Question 12
    Use Newton's method with an initial approximation \( x_1 = 2 \) to find a second approximation to the solution of the equation \( e^x = x^3 \).
    Show Solution to Question 12

    Let \( f(x) = e^x - x^3 = 0 \). Then \( f'(x) = e^x - 3x^2 \).

    Using Newton's algorithm \( x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)} \) with \( x_1 = 2 \):

    \[ x_2 = 2 - \frac{e^2 - 2^3}{e^2 - 3(2^2)} \approx 1.87 \]
13. Question 13
    Find the absolute maximum and minimum of the function \( f(x) = x^4 - x^3 \) on the interval \( [0, 5] \).
    Show Solution to Question 13

    Derivative: \( f'(x) = 4x^3 - 3x^2 = x^2(4x - 3) \). Critical points are \( x = 0 \) and \( x = 3/4 \).

    Evaluating at endpoints and critical points on \( [0, 5] \):

    \[ f(0) = 0, \quad f(5) = 500, \quad f(3/4) = -\frac{27}{256} \]

    Absolute maximum is \( 500 \) at \( x = 5 \), and absolute minimum is \( -\dfrac{27}{256} \) at \( x = 3/4 \).

14. Question 14
    If the dimensions \( L \), \( W \), and \( H \) of a rectangular box are changing at the rates \( \dfrac{dL}{dt} = 0.1 \; \text{cm/sec} \), \( \dfrac{dW}{dt} = -0.2 \; \text{cm/sec} \), and \( \dfrac{dH}{dt} = 0.3 \; \text{cm/sec} \), at what rate is the volume of the box changing when \( L = 20 \; \text{cm} \), \( W = 8 \; \text{cm} \), and \( H = 5 \; \text{cm} \)?
    Show Solution to Question 14

    Volume formula: \( V = LWH \). Differentiating with respect to time \( t \):

    \[ \frac{dV}{dt} = WH\frac{dL}{dt} + LH\frac{dW}{dt} + LW\frac{dH}{dt} \]

    Substituting given values (\( L = 20, W = 8, H = 5, \dfrac{dL}{dt} = 0.1, \dfrac{dW}{dt} = -0.2, \dfrac{dH}{dt} = 0.3 \)):

    \[ \frac{dV}{dt} = (8)(5)(0.1) + (20)(5)(-0.2) + (20)(8)(0.3) = 32 \; \text{cm}^3/\text{sec} \]
15. Question 15
    What are the dimensions of the rectangle with the largest area that can be inscribed in a semicircle of radius 3?
    

    

    Show Solution to Question 15

    Circle equation: \( x^2 + y^2 = 9 \implies y = \sqrt{9 - x^2} \). Rectangle dimensions: \( L = 2x \) and \( W = \sqrt{9 - x^2} \).

    Area function: \( A(x) = 2x\sqrt{9 - x^2} \) for \( 0 \le x \le 3 \).

    

    Derivative: \( A'(x) = \dfrac{2(-2x^2 + 9)}{\sqrt{9 - x^2}} \). Setting \( A'(x) = 0 \) gives critical point \( x = \sqrt{4.5} \).

    Evaluating gives maximum area at \( x = \sqrt{4.5} \), leading to dimensions:

    \[ L = 2\sqrt{4.5} \approx 4.24, \quad W = \sqrt{9 - 4.5} \approx 2.12 \]

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