# Calculus 1 Practice Question - A

A set of Calculus 1 questions with their detailed solutions to practice for tests, exams, placement exams, ... and gain deep understanding on the following topics:
1. Functions
2. Limits
3. Continuity
4. Derivatives
5. Applications of Derivatives
The questions were designed to cover the most important topics in calculus 1 and their detailed solutions include links to more practice in these topics.

1. Question 1
Find the domain of function $$f(x) = \dfrac{\sqrt{x - 1}}{\sqrt{4 - x^2}}$$.

2. Question 2
Find the range of function $$f(x) = \dfrac{x - 1}{2-3x}$$.    

3. Question 3
Find the inverse of function $$f(x) = ln (2x - 3) + 2$$.

4. Question 4
Evaluate each of the following limits
a) $$\lim_{x\to 16} \dfrac{-\dfrac{1}{\sqrt x} + \dfrac{1}{4}}{x - 16 }$$.

b) $$\lim_{x\to +\infty} \dfrac{-x^3+2x-1}{x^4 - 3 x^3 + 9 }$$.

c) $$\lim_{x\to +\infty} x \sin(\dfrac{3}{x})$$.

d) $$\lim_{x\to 0} \dfrac{sin(x)+x}{2x^2+x}$$.

e) $$\lim_{x\to +\infty} \dfrac{sin(x)+1}{x}$$.

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

6. Question 6
Find the derivative of the following functions and do not simplify the final answer.

a) $$f(x) = e^{x-1} + \ln (3x-1) + \sin(2x+1)$$

b) $$g(x) = (2x-1)^2(\tan(x)-1)$$

c) $$h(x) = \dfrac{x - \cos(x)}{x^2-2x+1}$$

d) $$m(x) = \sin \left(\sqrt{x^3 - \dfrac{1}{x} + 2} \right)$$

e) $$n(x) = 3^{ 2x+3} + \log_3(2x-1)$$

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

8. Question 8
Find the constants $$a$$ and $$b$$ so that the function $$f$$ is continuous on $$(-\infty , +\infty )$$
$$f(x) = \left\{ \begin{array}{ll} 2x - 1 & x\le 1 \\ a x^3 + b & 1 \lt x \lt 2 \\ x + 2 b & x\ge 2 \\ \end{array} \right.$$

9. Question 9
Find the equation of the tangent line to the curve with equation $$y = x + \sin(x)$$ at $$x = 0$$.

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

11. Question 11
Determine on what interval(s) is the function $$f(x) = e^x(x^2-5x+8)+\dfrac{x^4}{12}-\dfrac{x^3}{6}$$ concave up and concave down and any point of inflection.

12. Question 12
Use Newton's method with initial approximation $$x_1 = 2$$ to find a second approximation to the solution of the equation $$e^x = x^3$$.

13. Question 13
Find the absolute maximum and minimum of the function $$f(x) = x^4 - x^3$$ on the interval $$[0,5]$$.

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

15. Question 15
What are the dimensions of the rectangle with the largest area that can be inscribed in a semicircle of radius 3? ## Detailed Solutions to the Above Questions

1. 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; hence the inequalities to solve
$$x - 1 \ge 0$$ and $$4 - x^2 \gt 0$$
The solution sets for the first and second inequalities are respectively
$$x \ge 1$$ and $$-2 \lt x \lt 2$$
Both inequalities must be satisfied simultaneously, therefore the domain of the given function is the intersection of the sets $$x \ge 1$$ and $$-2 \lt x \lt 2$$ which is given by
$$1 \le x \lt 2$$

2. Solution to Question 2
According to the properties of inverse fucntions, one way to find the range of the given function is to find the domain of its inverse.
Let us first proove that $$f$$ is a one to one function and then find its inverse. To proove that $$f$$ is a one to one function and hence invertible, we use the contrapositive of the one to one function and start with $$f(a) = f(b)$$ and proove that $$a = b$$. Hence
$$\dfrac{a - 1}{2-3 a} = \dfrac{b - 1}{2-3 b}$$
Cross multiply
$$(a - 1)(2-3 b) = (2 - 3 a)(b - 1)$$
Expand
$$2a - 2 - 3 a b + 3 b = 2 b - 2 - 3 ab + 3 a$$
Group like terms
$$2 a = 2 b$$
Solve for a
$$a = b$$
which prooves that function $$f$$ is a one to one function and therefore has an inverse
The inverse of $$f$$ may be calculated starting with the equation
$$y = \dfrac{x - 1}{2-3x}$$
Cross multiply the above equation
$$2 y - 3 x y = x - 1$$
and solve for $$x$$
$$x = \dfrac{2 y + 1}{3y + 1}$$
Interchange $$x$$ and $$y$$ in the above equation to obtain the inverse $$f^{-1}$$
$$f^{-1}(x) = y = \dfrac{2 x + 1 }{3x + 1}$$.
The domian of $$f^{-1}$$ is the set of all real numbers except $$-\dfrac{1}{3}$$. Hence the range of $$f$$ is the set of all real numbers except $$-\dfrac{1}{3}$$ which may be written in interval form as
$$(-\infty , - \dfrac{1}{3}) \cup (- \dfrac{1}{3} , +\infty)$$

3. Solution to Question 3
Write the function as an equation as follows
$$y = ln (2x - 3) + 2$$
Solve the above for x
$$y - 2 = ln (2x - 3)$$
$$2x - 3 = e^{y - 2}$$
$$2 x = e^{y - 2} + 3$$
$$x = \dfrac{1}{2} (e^{y - 2} + 3)$$
Interchange $$x$$ and $$y$$
$$y = \dfrac{1}{2} (e^{x - 2} + 3)$$
The inverse of $$f$$ is given by
$$f^{-1}(x) = \dfrac{1}{2} (e^{x - 2} + 3)$$

4. 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 $$-\dfrac{1}{\sqrt x} - \dfrac{1}{4}$$

$$\lim_{x\to 16} \dfrac{-\dfrac{1}{\sqrt x} + \dfrac{1}{4}}{x - 16 }$$ = $$\lim_{x\to 16} \dfrac{ (-\dfrac{1}{\sqrt x} + \dfrac{1}{4})(-\dfrac{1}{\sqrt x} - \dfrac{1}{4}) }{ (x - 16)(-\dfrac{1}{\sqrt x} - \dfrac{1}{4}) }$$
Simplify
$$= \lim_{x\to 16} \dfrac{ \dfrac{1}{x} - \dfrac{1}{16} }{ (x - 16)(-\dfrac{1}{\sqrt x} - \dfrac{1}{4}) } = \lim_{x\to 16} \dfrac{ \dfrac{16-x}{16x} }{ (x - 16)(-\dfrac{1}{\sqrt x} - \dfrac{1}{4}) }$$

$$= \lim_{x\to 16} \dfrac{-1}{16x(-\dfrac{1}{\sqrt x} - \dfrac{1}{4})} = \dfrac{-1}{16 \times 16(-\dfrac{1}{\sqrt 16} - \dfrac{1}{4})} = \dfrac{1}{128}$$

b)
The limit is of the indeterminate form $$\dfrac{\infty}{\infty}$$.
Divide all terms in the numerator and all terms of the denominator by with the term with the highest power which in $$x^4$$
$$\lim_{x\to +\infty} \dfrac{-x^3+2x-1}{x^4 - 3 x^3 + 9 } = \lim_{x\to +\infty} \dfrac{\dfrac{-x^3}{x^4}+\dfrac{2x}{x^4}-\dfrac{1}{x^4}}{\dfrac{x^4}{x^4} - \dfrac{3 x^3}{x^4} + \dfrac{9}{x^4} }$$.
Simplify rational terms
$$= \lim_{x\to +\infty} \dfrac{\dfrac{-1}{x}+\dfrac{2}{x^3}-\dfrac{1}{x^4}}{1 - \dfrac{3}{x} + \dfrac{9}{x^4} } = \dfrac{0+0-0}{1 - 0 + 0 } = \dfrac{0}{1} = 0$$

c)
The limit is of the indeterminate form $$\infty \cdot 0$$.
Let $$t = \dfrac{3}{x}$$ and rewrite the limit in terms of t.
$$\lim_{x\to +\infty} x \sin(\dfrac{3}{x}) = \lim_{t\to 0} 3 \dfrac {\sin(t)}{t}$$.
Using the well known result $$\lim_{t\to 0} \dfrac {\sin(t)}{t} = 1$$, the limit evaluates to
$$= 3 \times 1 = 3$$

d)
$$\lim_{x\to 0} \dfrac{\sin(x)+x}{2x^2+x} = \dfrac{0}{0}$$, indeterminate form
Use the a L'Hospital's Rule to the indeterminate form, we can write
$$\lim_{x\to 0} \dfrac{\sin(x)+x}{2x^2+x} = \lim_{x\to 0} \dfrac{ d (\sin(x)+x) / dx }{ d(2x^2+x)/dx} = \lim_{x\to 0} \dfrac{ cos(x)+ 1 }{ 4x+1} = \dfrac{ cos(0)+ 1 }{ 4\times 0+1} = 2$$

e)
$$\lim_{x\to +\infty} \dfrac{\sin(x)+1}{x}$$.
It is well known that the range of $$\sin(x)$$ is given by
$$-1 \le \sin(x) \le 1$$
Add 1 to all terms of the inequality to obtain the following inequality
$$-1 + 1 \le \sin(x) + 1 \le 1 + 1$$
$$0 \le \sin(x) + 1 \le 2$$
Divide all terms of the above inequality by positive $$x$$
$$\dfrac{0}{x} \le \dfrac{\sin(x) + 1}{x} \le \dfrac{2}{x}$$
We have $$\lim_{x\to +\infty} \dfrac{0}{x} = 0$$ and $$\lim_{x\to +\infty} \dfrac{2}{x} = 0$$
Using the squeezing (or Sandwich) theorem, we can evaluate the given limit as follows
$$\lim_{x\to +\infty} \dfrac{\sin(x)+1}{x} = 0$$

5. Solution to Question 5
a)
The graph of $$y = e^{x-1}$$ and $$y = x$$ are shown below. The derivative of $$y = e^{x-1}$$ is equal to $$y' = e^{x-1}$$ and the slope $$m$$ of the tangent at $$x = 1$$ is the value of the derivative at $$x = 1$$. Hence $$m = e^{1-1}= 1$$
The equation of the tangent line at the point $$(1,1)$$ is given by
$$y - 1 = 1 \times (x - 1)$$
Which simplifies to
$$y = x$$
Hence the graphs of $$y = e^{x-1}$$ and $$y = x$$ are tangent at the point $$(1,1)$$ and we can therefore state graphically that $$e^{x-1} \ge x$$.
b)
The first and secopnd derivatives of $$f$$ are given by
$$f'(x) = \dfrac{x^2}{2}-e^{x-1}$$
$$f''(x) = x-e^{x-1}$$
A point of inflection occurs at a value of $$x$$ where $$f''(x)$$ change sign. We have seen in part a) that $$e^{x-1} \ge x$$ which may be written as
$$x-e^{x-1} \le 0$$
and therefore $$f''(x)$$ is negative and has a zero at $$x = 1$$. Hence the graph of $$f(x)$$ is concave down and does not have a point of inflection because $$f''(x)$$ does not change sign.

6. Solution to Question 6
a) Sum rule of derivatives gives: $$f'(x) = e^{x-1} + \dfrac{3}{3x-1} + 2 \cos(2x+1)$$
b) Product of derivatives rule: $$g'(x) = 4(2x-1)(\tan(x)-1) + (2x-1)^2(\sec^2(x))$$
c) quotient of derivatives rule: $$h'(x) = \dfrac{ (1 + \sin(x))(x^2-2x+1) - (x - \cos(x))(2x-2) }{(x^2-2x+1)^2}$$

d) Let $$u = \sqrt{x^3 - \dfrac{1}{x} + 2}$$ , write function $$m$$ as $$m = \sin u$$ then use the chain rule of derivatives

$$m'(x) = \dfrac{d m}{d u} \dfrac{d u}{d x} = \cos(u) \dfrac{1}{2}(3x^2+\dfrac{1}{x^2})(x^3 - \dfrac{1}{x} + 2)^{-1/2}$$

$$= \dfrac{1}{2} \cos \left(\sqrt{x^3 - \dfrac{1}{x} + 2} \right) (3x^2+\dfrac{1}{x^2})(x^3 - \dfrac{1}{x} + 2)^{-1/2}$$

e)
Rewrite $$3^{ 2x+3}$$ and $$\log_3(2x-1)$$ as
$$3^{2x+3} = e^{(2x+3) \ln 3}$$ , change of base of exponentials
$$\log_3(2x-1) = \dfrac{ \ln(2x-1)}{ \ln 3}$$ , change of base of logarithms
Substitute and rewrite $$n(x)$$ as
$$n(x) = 3^{ 2x+3} + \log_3(2x-1) = e^{(2x+3) \ln 3} + \dfrac{ \ln(2x-1)}{ \ln 3}$$
We now calculate the derivative
$$n'(x) = ( 2 \ln 3 ) e^{(2x+3) \ln 3} + \dfrac{1}{ \ln 3} \dfrac{2}{2x-1} = ( 2 \ln 3 ) 3^{2x+3} + \dfrac{2}{\ln 3(2x-1)}$$

7. Solution to Question 7
We first differentiate the given equation implicitly
$$2 y \dfrac{d y}{d x} cos(y^2) = 2 x$$
$$\dfrac{d y}{d x} = \dfrac{ x}{ y cos(y^2)}$$
The slope $$m$$ of the tangent line is given by the value of $$\dfrac{d y}{d x}$$ at the point $$( 0,\sqrt{\pi})$$.
$$m = \dfrac{(0)}{\sqrt{\pi} cos((\sqrt{{\pi}})^2)} = 0$$
The equation, in point slope form, of the tangent line to the curve at the point $$( 0,\sqrt{\pi})$$ is given by
$$y - \sqrt{\pi} = 0(x - 0)$$
It is a horizontal line given by
$$y = \sqrt{\pi}$$

8. Solution to Question 8
$$f(x)$$ is continuous on the intervals $$(-\infty , 1)$$ , $$(1,2)$$ and $$(2 , +\infty)$$. We need to find $$a$$ and $$b$$ so that it is also continous at $$x = 1$$ and $$x = 2$$ and therefore continuous on $$(-\infty , +\infty )$$.
$$f(1) = 1$$
$$\lim_{x\to 1^-} f(x) = 1$$
$$\lim_{x\to 1^+} f(x) = a(1)^3+b = a + b$$
The limits from the left and right of $$1$$ must be equal
$$a + b = 1$$ (equation 1)
$$f(2) = 2 + 2 b$$
$$\lim_{x\to 2^-} f(x) = a(2)^3 + b = 8 a + b$$
$$\lim_{x\to 2^+} f(x) = 2 + 2 b$$
The limits from the left and right of $$2$$ must be equal
$$8 a + b = 2 + 2 b$$ (equation 2)
Solve equations (1) and (2) simultaneously to find
$$a = \dfrac{1}{3}$$ and $$b = \dfrac{2}{3}$$

9. Solution to Question 9
Find the derivative of $$y$$.
$$y' = 1 + \cos(x)$$
The slope $$m$$ of the tangent at $$x = 0$$ is equal to the value of $$y'$$ at $$x = 0$$. Hence
$$m = 1 + \cos(0) = 2$$
The y coordinate of the point of tangency $$P$$ is given by the value of $$y$$ at $$x = 0$$. Hence
$$P(0 , 0 + \sin(0)) = P(0,0)$$
The equation of the tangent line in point slope form is given by
$$y - 0 = 2(x - 0)$$
and in slope intercept form
$$y = 2 x$$

10. Solution to Question 10
Definition of the derivative $$f'$$ of function $$f$$ is given by the limit
$$f'(x) = \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$
Substitute $$f(x)$$ by $$\sqrt{x+2}$$ in the above definition to obtain
$$f'(x) = \lim_{h\to 0} \dfrac{\sqrt{x + h +2}- \sqrt{x+2} }{h}$$
The above limit is of the indeterminate form $$\dfrac{0}{0}$$. Multiply the numerator and denominator by the conjugate of the numerator
$$f'(x) = \lim_{h\to 0} \dfrac{ (\sqrt{x + h +2}- \sqrt{x+2} ) (\sqrt{x + h +2} + \sqrt{x+2} ) }{h(\sqrt{x + h +2} + \sqrt{x+2} )}$$
Expand the numerator and simplify
$$f'(x) = \lim_{h\to 0} \dfrac{ (x + h + 2)- (x + 2) ) }{h(\sqrt{x + h +2} + \sqrt{x+2} )} = \lim_{h\to 0} \dfrac{ h }{h(\sqrt{x + h +2} + \sqrt{x+2} )}$$
Divide numerator and denominator by $$h$$ (or cancel $$h$$)
$$f'(x) = \lim_{h\to 0} \dfrac{ 1 }{\sqrt{x + h +2} + \sqrt{x+2} }$$
Evaluate the limit and hence the derivative
$$f'(x) = \dfrac{ 1 }{ \sqrt{x + 0 + 2} + \sqrt{x + 2} } = \dfrac{ 1 }{2 \sqrt{x + 2} }$$

11. Solution to Question 11
Find the first and second derivatives
$$f'(x) = e^x (x^2-5x+8) + e^x (2 x -5) +\dfrac{x^3}{3}-\dfrac{x^2}{2} = e^x (x^2-3x+3)+\dfrac{x^3}{3}-\dfrac{x^2}{2}$$
$$f''(x) = e^x(x^2-3x+3)+e^x(2x-3) + x^2-x = e^x(x^2-x)+x^2-x = x (x - 1) e^x$$
$$f''$$ has two zeros: $$x = 0$$ and $$x = 1$$ and $$e^x$$ is always positive. Hence the table of signs of $$f''$$ has three intervals
1) $$(-\infty , 0 )$$ , test value $$x = -1$$ , $$f''(-1) = 2/e$$ , hence $$f''(x)$$ is positive on the interval $$(-\infty , 0 )$$.
2) $$(0 , 1 )$$ , test value $$x = 1/2$$ , $$f''(1/2) = -\dfrac{\sqrt 2}{4}$$ , hence $$f''(x)$$ is negative on the interval $$(0 , 1 )$$.
3) $$(1 , +\infty )$$ , test value $$x = 2$$ , $$f''(2) = 2e^2$$ , $$f''(x)$$ is positive on the interval $$(1 , +\infty )$$.
$$f''$$ is concave up on the intervals $$(-\infty , 0 )$$ and $$(1 , +\infty )$$ , and concave down on the interval $$(0 , 1 )$$.
$$f''$$ changes sign at $$x = 0$$ and $$x = 1$$ and therefore has points of inflection at $$x = 0$$ and $$x = 1$$.

12. Solution to Question 12
Newton's method is based on the following algorithm: knowing an approximation $$x_n$$ to the solution of an equation $$f(x) = 0$$, the next approximation $$x_{n+1}$$ to the equation is given by
$$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$$.
The solution to the given equation $$e^x = x^3$$ is equal to the solution of the equation $$f(x) = e^x - x^3 = 0$$
$$f'(x) = e^x - 3x^2$$
We know a first approximation $$x_1 = 2$$; using Newton's algorithm we approximate $$x_2$$ by
$$x_{2} = x_1 - \dfrac{f(x_1)}{f'(x_1)} = 2 - \dfrac{e^2 - 2^3}{e^2 - 3 \times 2^2} \approx 1.87$$ (rounded to two decimal places)
An approximation $$x_3$$ may be obtained using $$x_2$$ found above and so on.

13. Solution to Question 13
Find the first derivative and factor it.
$$f'(x) = 4 x^3 - 3 x^2 = x^2(4x - 3)$$
$$f'(x)$$ has two zeros at $$x = 0$$ and $$x = 3/4$$ and both are within the interval $$[0,5]$$. The zeros of $$f'(x)$$ are called critical points.
We now evalute the function at the endpoints of the given interval and the zeros of $$f'(x)$$.
$$f(0) = 0$$
$$f(5) = 5^4 - 5^3 = 500$$
$$f(3/4) = (3/4)^4 - (3/4)^3 = -\dfrac{27}{256}$$
Comparing these values, $$f(x)$$ has an absolute maximum equal to $$500$$ at the endpoint $$x = 5$$ and an absolute minimum equal to $$-\dfrac{27}{256}$$ at the critical point $$x = 3/4$$
More questions on absolute minimum and maximum of a function included.

14. Solution to Question 14
The volume $$V$$ of rectangular box with dimensions $$L$$,$$W$$ and $$H$$ is given by
$$V = L(t) W(t) H(t)$$
where $$L(t)$$ , $$W(t)$$ and $$H(t)$$ are functions of time $$t$$.
Since the dimensions of the box are changing with time, the volume of the box is also changing with time at the rate of change of $$V$$ is given by the first derivative $$\dfrac{dV}{dt}$$.
$$\dfrac{dV}{dt} = W(t) H(t) \dfrac{d L}{dt } + L(t) H(t) \dfrac{d W}{dt } + L(t) W(t) \dfrac{d H}{dt }$$
Substitute the known quantities by their numerical values
$$\dfrac{dV}{dt} = 8 \times 5 \times 0.1 + 20 \times 5 \times (-0.2) + 20 \times 8 \times 0.3 = 32 \; cm^3 / sec$$

15. Solution to Question 15
Equation of a circle with radius $$3$$ and center at $$(0 , 0)$$given by
$$x^2 + y^2 = 3^2$$
Solve the above equation fo y
$$y^2 = 9 - x^2$$
$$y = \pm \sqrt{9 - x^2}$$
The equation for the upper semi cricle is $$y = \sqrt(9-x^2)$$
A point on the semi circle with x coordinate $$x$$ has a y coordinate equal to $$\sqrt{9 - x^2}$$ (see graph below)
The rectangle has a length $$L = 2x$$ and a width (or height) $$W = \sqrt(9-x^2)$$. The area $$A$$ of the rectangle is given by
$$A(x) = L \times W = 2 x \sqrt(9-x^2)$$ , $$0 \le x \le 3$$ Find first derivative of $$A$$
$$\dfrac{d A}{dx} = 2\sqrt{9-x^2} + (2x) (\dfrac{1}{2}) (-2x) (9-x^2)^{-1/2} =\dfrac{2\left(-2x^2+9\right)}{\sqrt{9-x^2}}$$
Zeros of $$\dfrac{d A}{dx}$$ are given by the zeros of the numerator of $$\dfrac{d A}{dx}$$
$$-2x^2+9 = 0$$
Gives two zeros: $$x = \sqrt{4.5}$$ and $$x = - \sqrt{4.5}$$
We consider the zero at $$x = \sqrt{4.5}$$ since it is within the interval $$[0 , 3 ]$$
We now evaluate the area $$A(x)$$ at the endpoints $$x = 0$$ and $$x = 3$$ and at the critical point $$x = \sqrt{4.5}$$.
$$A(0) = 0$$
$$A(3) = 0$$
$$A(\sqrt{4.5}) = 9$$
The area A is maximum is for $$x = \sqrt{4.5}$$
The dimensions are:
$$L = 2 x = 2 \sqrt{4.5} \approx = 4.24$$
$$W = \sqrt(9-x^2) = \sqrt(9-\sqrt{4.5}^2) \approx 2.12$$
More optimization problems included.