QUESTION IMAGE
Question
graph the following function. state the domain and range. is the function increasing or decreasing?
$f(x) = \sqrt3{x - 8}$
choose the correct graph below
\\(\bigcirc\\) a.
\\(\bigcirc\\) b.
\\(\bigcirc\\) c.
\\(\bigcirc\\) d.
Step 1: Analyze the function \( f(x)=\sqrt[3]{x - 8} \)
The cube - root function \( y = \sqrt[3]{u}\) has a domain of all real numbers (\(u\in\mathbb{R}\)) and a range of all real numbers (\(y\in\mathbb{R}\)). For the function \( f(x)=\sqrt[3]{x - 8}\), if we let \( u=x - 8\), then the domain of \( f(x)\) is all real numbers because we can take the cube - root of any real number. To find the domain, we solve for \( x\) in the expression inside the cube - root. Since there are no restrictions (unlike square - root functions where the expression inside must be non - negative), the domain of \( f(x)=\sqrt[3]{x - 8}\) is \( (-\infty,\infty)\) (all real numbers). The range of a cube - root function \( y=\sqrt[3]{x - 8}\) is also all real numbers because as \( x\) takes on all real values, \( x - 8\) takes on all real values, and the cube - root of all real values is also all real values.
Step 2: Analyze the end - behavior and monotonicity
The derivative of the function \( y = \sqrt[3]{x-8}=(x - 8)^{\frac{1}{3}}\) using the power rule \(\frac{d}{dx}(x^n)=nx^{n - 1}\) is \(y^\prime=\frac{1}{3}(x - 8)^{-\frac{2}{3}}=\frac{1}{3(x - 8)^{\frac{2}{3}}}=\frac{1}{3\sqrt[3]{(x - 8)^2}}\). The derivative \(y^\prime\) is positive for all \(x
eq8\) (because the denominator \(3\sqrt[3]{(x - 8)^2}\) is always positive for \(x
eq8\), and the numerator is 1). At \(x = 8\), the function is continuous (since the limit as \(x
ightarrow8\) of \(\sqrt[3]{x - 8}\) is 0 and \(f(8)=\sqrt[3]{8 - 8}=0\)) and the derivative does not exist (it has a vertical tangent at \(x = 8\)). So the function \(f(x)=\sqrt[3]{x - 8}\) is an increasing function for all real numbers \(x\).
Step 3: Analyze the graph of the function
The parent function of \(y=\sqrt[3]{x}\) has a point at \((0,0)\) and passes through \((1,1)\) and \((- 1,-1)\). The function \(y=\sqrt[3]{x - 8}\) is a horizontal shift of the parent function \(y = \sqrt[3]{x}\) to the right by 8 units. So the graph of \(y=\sqrt[3]{x - 8}\) passes through the point \((8,0)\) (when \(x = 8\), \(y=\sqrt[3]{8 - 8}=0\)). As \(x
ightarrow\infty\), \(y=\sqrt[3]{x - 8}
ightarrow\infty\) and as \(x
ightarrow-\infty\), \(y=\sqrt[3]{x - 8}
ightarrow-\infty\). The function is increasing, so as \(x\) increases, \(y\) increases.
Now, let's analyze the options:
- Option A: The graph should be a shifted cube - root graph. The cube - root graph has a shape that passes through the point corresponding to the shift. Since our function is \(y=\sqrt[3]{x - 8}\), the key point is \((8,0)\). The graph of an increasing cube - root function shifted right by 8 units will have the correct shape. The function is increasing, so as \(x\) increases, \(y\) increases.
- Option B: If the graph is not increasing (or has the wrong shape for a cube - root function), it can be eliminated.
- Option C: The shape of the graph does not match the shape of a cube - root function. A cube - root function has a more "pointed" shape near the point of the shift and is symmetric in a different way than a function that might be misrepresented here.
- Option D: The graph does not have the correct shape for an increasing cube - root function shifted right by 8 units.
Step 4: State the domain, range, and monotonicity
- Domain: The domain of \(f(x)=\sqrt[3]{x - 8}\) is all real numbers, so the domain is \((-\infty,\infty)\) (or \(\mathbb{R}\)).
- Range: The range of \(f(x)=\sqrt[3]{x - 8}\) is all real numbers, so the range is \((-\infty,\infty)\) (or \(\mathbb{R}\)).
- Monotonicity: The function \(f(x)=\sqrt[3]{x - 8}\) is an increasing function because its…
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- Correct graph: Option A (assuming the visual of Option A matches the graph of \(y = \sqrt[3]{x-8}\) which is an increasing cube - root function shifted right by 8 units).
- Domain: \((-\infty,\infty)\)
- Range: \((-\infty,\infty)\)
- The function is increasing.