Math Problem Statement
Carefully graph f(x) = {√x ; 0 ≤ x < 4, (x - 6)^2 - 2 ; 4 ≤ x ≤ 10} below. Then, write a detailed slope statement that describes the changes in slope over the interval.
Solution
The given piecewise function is defined as:
\begin{cases} \sqrt{x}, & 0 \leq x < 4 \\ (x - 6)^2 - 2, & 4 \leq x \leq 10 \end{cases}$$ ### Step 1: Graphing the function #### For $$0 \leq x < 4$$, the function is $$f(x) = \sqrt{x}$$: - This is the graph of a square root function. - At $$x = 0$$, $$f(x) = \sqrt{0} = 0$$. - At $$x = 4$$, $$f(x) = \sqrt{4} = 2$$, but we stop just before $$x = 4$$. So, this part of the graph is the square root curve, starting from (0, 0) and ending near (4, 2) but not including the point (4, 2). #### For $$4 \leq x \leq 10$$, the function is $$f(x) = (x - 6)^2 - 2$$: - This is a quadratic function (a parabola) that opens upwards, shifted to the right by 6 units and down by 2 units. - At $$x = 4$$, $$f(4) = (4 - 6)^2 - 2 = (-2)^2 - 2 = 4 - 2 = 2$$, so the parabola starts from $$(4, 2)$$. - At $$x = 6$$, $$f(6) = (6 - 6)^2 - 2 = 0 - 2 = -2$$, so the vertex is at $$(6, -2)$$. - At $$x = 10$$, $$f(10) = (10 - 6)^2 - 2 = 16 - 2 = 14$$, so the parabola extends to $$(10, 14)$$. ### Step 2: Describing the Slope Changes - **For $$0 \leq x < 4$$**: - The slope of $$f(x) = \sqrt{x}$$ is positive but decreasing as $$x$$ increases. The function is concave down, meaning that as you move to the right, the slope of the tangent line decreases (it becomes flatter). - The slope approaches zero as $$x$$ approaches 4 from the left, making the graph less steep. - **At $$x = 4$$**: - The function switches from $$f(x) = \sqrt{x}$$ to the quadratic function $$(x - 6)^2 - 2$$. - The value of the function at $$x = 4$$ is continuous (both pieces meet at $$(4, 2)$$), but the slopes are not continuous. The slope of $$\sqrt{x}$$ approaches zero from the left, but the slope of the quadratic function is positive at $$x = 4$$. - **For $$4 \leq x \leq 10$$**: - The slope is initially negative, as the parabola is decreasing until the vertex at $$x = 6$$. - At $$x = 6$$, the slope is zero (the minimum point or vertex). - For $$x > 6$$, the slope becomes positive again, increasing as $$x$$ moves toward 10. The parabola is concave up, so the slope increases rapidly as you move away from the vertex. Would you like a detailed sketch of this graph or a more specific breakdown of the slope behavior? --- Here are five related questions to explore further: 1. What happens to the slope of a square root function as $$x \to 0^+$$? 2. How do you find the derivative of a piecewise function at the boundary point (in this case, at $$x = 4$$)? 3. How do you interpret concavity in relation to the sign of the second derivative for each piece of the function? 4. What effect does shifting a quadratic function horizontally and vertically have on its graph and slope? 5. Can a piecewise function be differentiable at the boundary point? What conditions must be met? **Tip:** When analyzing piecewise functions, always check the continuity and differentiability at the points where the pieces join.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Square Root Function
Quadratic Functions
Slope Analysis
Graphing
Formulas
f(x) = √x
f(x) = (x - 6)^2 - 2
Theorems
Continuity of Piecewise Functions
Slope Behavior in Different Function Types
Concavity and Slope
Suitable Grade Level
Grades 10-12
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