📝 Explicit vs implicit functions (14 MCQs)

📖 From Calculus • 4. Topics in Differentiation • 14 questions available

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What is Explicit vs implicit functions?

Definition:
An explicit function directly expresses the dependent variable in terms of the independent variable, such as $y = f(x)$ . In contrast, an implicit function defines a relationship between variables where $y$ is not isolated, represented by an equation like $F(x, y) = 0$ , requiring special techniques to analyze the dependency between the variables involved.

Example:
Consider the circle equation $x^2 + y^2 = 25$ . This is implicit because $y$ is not isolated. To find $y$ explicitly, we solve for it: $y = \pm\sqrt{25 - x^2}$ , which gives two separate explicit functions for the upper and lower semicircles.

Reason:
Distinguishing these forms is crucial because explicit functions allow direct evaluation, while implicit relations often describe complex curves like circles or ellipses that cannot be represented by a single explicit function without splitting them into multiple parts.

Easy

Medium

Hard

📝 All Explicit vs implicit functions MCQs

Q1. Which of the following best describes an explicit definition of a function y in terms of x?

A.y appears alone on one side of the equation ✅

B.y appears with x on both sides

C.The equation cannot be solved for y

D.y is defined implicitly

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: An explicit definition requires that y be isolated on one side of the equation, showing y = f(x). This isolates y and makes the dependence on x clear, distinguishing it from implicit forms where y appears together with x on the same side.

Q2. Consider the equation $yx + y + 1 = x$ . Which statement is true about this equation?

A.It defines y explicitly as a function of x

B.It defines y implicitly as a function of x ✅

C.It does not define y as a function of x

D.It defines x as a function of y

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The given equation cannot be rearranged to isolate y directly, but algebraic manipulation yields $y = \frac{x-1}{x+1}$ . Hence y is defined as a function of x, but only after rewriting, making the definition implicit.

Q3. If an equation fails the vertical line test, what can be concluded about the functions it may define implicitly?

A.It cannot define any function of x

B.It may define one or more functions of x ✅

C.It defines a unique function of x

D.It defines only constant functions

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Failing the vertical line test means the whole curve is not a single‑valued function of x, yet portions of the curve can still be expressed as separate functions. Thus the equation may implicitly define multiple functions on different intervals.

Q4. Given the implicit equation $x^2 + y^2 = 1$ , solving for y yields $y = \pm\sqrt{1 - x^2}$ . What logical inference follows regarding the domain of each implicit function?

A.Both functions are defined for all real x

B.Both functions are defined only where

1 - x^2 \ge 0

✅

C.Only the positive branch is defined for

x \ge 0

D.The domain is unrestricted because the circle is closed

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The square‑root expression requires a non‑negative radicand, so the condition $1 - x^2 \ge 0$ must hold. This restricts x to the interval $[-1,1]$ for both the positive and negative branches, ensuring each defines a valid function on that domain.

Q5. For the equation $x = y^2$ , after solving for y we obtain $y = \pm\sqrt{x}$ . Which of the following statements correctly describes the relationship between the original equation and the two resulting functions?

A.The original equation is equivalent to the union of the two functions

B.The original equation defines a single function

C.Each function represents a separate portion of the original graph ✅

D.Both functions together fail the vertical line test

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The original relation describes a parabola opening to the right, which fails the vertical line test. Solving for y splits the curve into two separate branches—one positive, one negative—each of which individually passes the test and represents a distinct function.

Q6. Suppose an implicit equation $\sin(xy) = y$ has a solution $y = g(x)$ near $x = 0$ . Using the Implicit Function Theorem, which condition must be satisfied at $(0,0)$ for $g$ to exist?

\frac{\partial}{\partial y}\sin(xy) \neq 0

at (0,0) ✅

\frac{\partial}{\partial x}\sin(xy) \neq 0

at (0,0)

C.Both partial derivatives must be zero

D.No condition is needed

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The Implicit Function Theorem requires the partial derivative with respect to y of the left‑hand side, evaluated at the point, to be non‑zero. At (0,0), $\partial_y\sin(xy)=\cos(0)\cdot x = 0$ unless x≠0; thus the theorem demands a non‑zero value, ensuring a locally unique function exists.

Q7. Which of the following equations can be solved explicitly for y without involving radicals or implicit functions?

yx + y + 1 = x

✅

x^2 + y^2 = 1

x = y^2

x^3 + y^3 = 3xy

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The first equation can be rearranged algebraically to isolate y, yielding $y = \frac{x-1}{x+1}$ . The other equations either involve radicals after solving (circle, parabola) or lead to higher‑degree polynomials that cannot be expressed in a simple explicit form.

Q8. Compare the number of branches (distinct functions) defined implicitly by the equations $x = y^2$ and $x^2 + y^2 = 1$ . Which statement is correct?

A.Both define exactly two branches ✅

B.The first defines two branches, the second defines one

C.The first defines one branch, the second defines two

D.Both define a single branch

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Solving $x = y^2$ gives $y = \pm\sqrt{x}$ , two branches. Solving the circle equation yields $y = \pm\sqrt{1 - x^2}$ , also two branches (upper and lower semicircles). Hence each implicit equation leads to exactly two distinct functions.

Q9. If we square both sides of the implicit relation $y = \sqrt{x}$ to obtain $y^2 = x$ , what effect does this have on the set of points represented?

A.It adds extraneous points where

y < 0

✅

B.It removes points where

y > 0

C.It leaves the set unchanged

D.It doubles the x-values

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Squaring eliminates the sign information of y, so the relation $y^2 = x$ includes points with both positive and negative y that satisfy the equation, whereas the original relation only allowed non‑negative y. Thus extra points with $y < 0$ are introduced.

Q10. Consider the implicit function defined by $x^3 + y^3 = 3xy$ . Which of the following statements about its differentiability at the point $(1,1)$ is true?

A.The function is not differentiable because the gradient is zero

B.The function is differentiable and

dy/dx = 1

✅

C.The function is differentiable and

dy/dx = -1

D.Differentiability cannot be determined without solving for y

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Differentiating implicitly gives $3x^2 + 3y^2 \frac{dy}{dx} = 3y + 3x \frac{dy}{dx}$ . Substituting $x=y=1$ simplifies to $3 + 3 \frac{dy}{dx} = 3 + 3 \frac{dy}{dx}$ , yielding a consistent equation and allowing the derivative to be computed as $dy/dx = 1$ . Hence the curve is smooth at (1,1).

Q11. Using implicit differentiation, find $\frac{dy}{dx}$ for the equation $yx + y + 1 = x$ . Which expression correctly represents the derivative?

\frac{dy}{dx} = \frac{1 - y}{x + 1}

✅

\frac{dy}{dx} = \frac{x - 1}{y + 1}

\frac{dy}{dx} = \frac{x - 1}{x + 1}

\frac{dy}{dx} = \frac{1 - x}{y + 1}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiating both sides with respect to x gives $y + x\frac{dy}{dx} + \frac{dy}{dx} = 1$ . Collecting $\frac{dy}{dx}$ terms yields $(x+1)\frac{dy}{dx} = 1 - y$ , so $\frac{dy}{dx} = \frac{1 - y}{x + 1}$ . This matches option A.

Q12. Why is implicit plotting particularly useful for visualizing the curve defined by $x^3 + y^3 = 3xy$ ?

A.Because the curve can be expressed as a simple explicit function

B.Because solving for y yields a single elementary formula

C.Because the curve consists of multiple branches that are hard to isolate explicitly ✅

D.Because the equation has no real solutions

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The Folium of Descartes does not lend itself to an elementary explicit expression for y; it produces several intertwined branches. Implicit plotting bypasses the need for solving for y, directly rendering the entire set of points that satisfy the equation, making it ideal for such complex curves.

Q13. Given the implicit relation $x^2 - y^2 = 4$ , determine whether it defines a function $y = f(x)$ over the interval $(-\infty, \infty)$ . Which conclusion is correct?

A.It defines a single-valued function for all real x

B.It defines two separate functions, each valid on a domain where

|x| \ge 2

✅

C.It does not define any function because it fails the vertical line test everywhere

D.It defines a piecewise function that is continuous for all x

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Solving for y gives $y = \pm\sqrt{x^2 - 4}$ . Real values exist only when $x^2 - 4 \ge 0$ , i.e., $|x| \ge 2$ . Thus two distinct functions exist on that domain, and no function exists for $|x|<2$ .

Q14. Explain how the vertical line test relates to the notion of implicit definition of functions. Which statement best captures this relationship?

A.If a curve passes the vertical line test, it cannot be defined implicitly

B.A curve that fails the vertical line test may still contain portions that are functions defined implicitly ✅

C.Implicit definitions always produce curves that pass the vertical line test

D.The vertical line test is irrelevant for implicit functions

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A curve may globally fail the vertical line test, meaning it is not a single function of x, yet subsets of the curve (branches) can satisfy the test and thus represent functions defined implicitly. Therefore, failure of the test does not preclude implicit functional portions.

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