Problem 46 - Calculus Single Variable

Question

Two functions \(f\) and \(g\) are given. Find constants \(h\) and \(k\) such that \(g(x)=f(x+h)+k .\) Describe the relationship between the plots of \(f\) and \(g\). \(f(x)=\sqrt{1-x^{2}}, g(x)=1+\sqrt{2 x-x^{2}}\)

Step-by-Step Solution

Answer

Constants are \(h = -1\) and \(k = 1\). Translate the graph of \(f(x)\) 1 unit right and shift 1 unit up for \(g(x)\).

Step 1: Analyze Given Functions

Let's examine the given functions. We have \( f(x) = \sqrt{1 - x^2} \) and \( g(x) = 1 + \sqrt{2x - x^2} \). Both functions involve square roots, suggesting they may represent semi-circles.

Step 2: Derive the Transformation Equation

The transformation equation given is \( g(x) = f(x + h) + k \). We need to adjust \( f(x) \) to match \( g(x) \) through horizontal shifts (determined by \( h \)) and vertical shifts (determined by \( k \)).

Step 3: Identify the Domain

The function \( f(x) = \sqrt{1 - x^2} \) has a domain \(-1 \leq x \leq 1\), while the expression under the square root in \( g(x) = \sqrt{2x - x^2} \) must be non-negative, which means solving \(2x - x^2 \geq 0\). Factoring gives \(x(2-x) \geq 0\). Thus, \(0 \leq x \leq 2\).

Step 4: Match the Two Functions

To match \(g(x)\) and \(f(x+h)+k\), observe that \( f(x+h) = \sqrt{1 - (x + h)^2} \) should equal \( \sqrt{2x - x^2} \) after transformation. Setting \( k \) for the vertical shift begins with \( k + \sqrt{1 - (x + h)^2} = 1 + \sqrt{2x - x^2} \).

Step 5: Solve for \(h\)

To solve for \(h\), recognize that the transformation is a perfect square comparison of semi-circle equations. By completing the square, we adjust domains by translating them. Since \(f\) and \(g\) shift along the x-axis, adjust their centers. Translate \(f\)'s center \((0,0)\) to \(g\)'s center \((1,0)\). Thus \( h = -1 \).

Step 6: Solve for \(k\)

For \(k\), shift the function vertically. Need \(f(1)\) when \(x = 0\) in \(g\). Since its maximum value for \(f(0) = 1 \), then \( k + f(-1) = 1 + \sqrt{2x - x^2} \) matches the transformation; obtain \(k = 1\). This fits \(g(1)=2\) implying \(k = 1\).

Step 7: Verify the Solutions

Substitute \(h = -1\) and \(k = 1\) back into the transformation equation. Check that \(f(x+h) + k\) computes \(g(x)\) directly by expanded checks: \(f(x-1)+1 = \sqrt{1-(x-1)^2}+1\) matches \(g(x)=1+\sqrt{2x-x^2}\). The transformation method holds.

Key Concepts

Horizontal Shift

A horizontal shift in function transformation involves moving a graph left or right on the x-axis. The principle behind this shift is adding or subtracting a constant value, denoted as \( h \), to the x-value in the original function. In the exercise, we determine the constant \( h \) that shifts the plot of the semi-circle function \( f(x) = \sqrt{1-x^2} \) so it corresponds with the g function.

• If \( h \) is positive, the function \( f(x+h) \) shifts left by \( h \) units.
• If \( h \) is negative, the function shifts right by \(|h|\) units.

In the solution provided, we see that the horizontal shift required for \( f(x) \) is \( h = -1 \). This means that the function \( f(x) \) is moved 1 unit to the right to match with \( g(x) = 1 + \sqrt{2x - x^{2}} \). This adjustment aligns the centers of the semicircles on the x-axis, transforming the center of \( f(x) \) from \((0, 0)\) to \((1, 0)\). This horizontal adjusting helps match the domain and appearance of \( g(x) \).

Vertical Shift

A vertical shift involves moving a graph up or down along the y-axis. The constant \( k \) in the transformation equation \( g(x) = f(x+h) + k \) dictates this shift. In the given exercise, the goal is to establish the precise value of \( k \) that vertically shifts \( f(x) = \sqrt{1-x^2} \) to form \( g(x) = 1 + \sqrt{2x - x^2} \).

• If \( k \) is positive, the function will move upward by \( k \) units.
• If \( k \) is negative, the function will move downward by \( |k| \) units.

For the transformation to occur, the vertical shift was determined as \( k = 1 \). This indicates that the graph of \( f(x) \) shifts 1 unit upwards, aligning the maximum value of \( f(x) \) with that of \( g(x) \). The vertical movement complements the horizontal adjustment, allowing for the accurate overlap of the two functions when their plots are compared.

Semi-Circle Functions

Semi-circle functions, such as the ones seen in this exercise, are classic forms defined by the equation \( f(x) = \sqrt{r^2 - x^2} \). These functions represent half of a circle and involve the principle of a square root, ensuring the expression under the root must be non-negative. The graph of a semi-circle function appears as the top half of the circle.

For \( f(x) = \sqrt{1 - x^2} \), its domain is defined by \(-1 \leq x \leq 1\), ensuring the value under the square root remains non-negative ending in a semi-circle with radius 1 centered at the origin. Similarly, the function \( g(x) = 1 + \sqrt{2x - x^2} \) is another semi-circle, but with a different expression under the radical.

In this exercised solution, recognizing each function's semi-circle form enabled easier comparison and determination of necessary shifts. It establishes the premise that both functions share the trait of representing semi-circles, hence understanding semi-circle equations became crucial. By comprehending these underlying geometric shapes, students can better visualize and solve transformations accurately.