Join IDNLearn.com today and start getting the answers you've been searching for. Ask your questions and get detailed, reliable answers from our community of experienced experts.
Sagot :
To solve for [tex]\((f \cdot g)(x)\)[/tex] given the functions [tex]\(f(x) = \sqrt{2x}\)[/tex] and [tex]\(g(x) = \sqrt{32x}\)[/tex], we proceed as follows:
1. Expression for [tex]\((f \cdot g)(x)\)[/tex]:
The notation [tex]\((f \cdot g)(x)\)[/tex] represents the product of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]. So, we need to multiply the expressions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f \cdot g)(x) = f(x) \cdot g(x) \][/tex]
2. Substitute the expressions for [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
Recall that [tex]\(f(x) = \sqrt{2x}\)[/tex] and [tex]\(g(x) = \sqrt{32x}\)[/tex]. Substituting these into the product:
[tex]\[ (f \cdot g)(x) = \sqrt{2x} \cdot \sqrt{32x} \][/tex]
3. Simplify the expression:
Using the property of square roots [tex]\(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\)[/tex], we combine the square roots:
[tex]\[ (f \cdot g)(x) = \sqrt{2x \cdot 32x} \][/tex]
4. Multiply the terms inside the square root:
Simplify [tex]\(2x \cdot 32x\)[/tex]:
[tex]\[ 2x \cdot 32x = 64x^2 \][/tex]
5. Take the square root of the resulting expression:
[tex]\[ \sqrt{64x^2} = \sqrt{64} \cdot \sqrt{x^2} = 8 \cdot x \][/tex]
Thus, the product function [tex]\((f \cdot g)(x)\)[/tex] simplifies to:
[tex]\[ (f \cdot g)(x) = 8x \][/tex]
To see this in action, let's plug in a specific value of [tex]\(x\)[/tex]:
- Let [tex]\(x = 4\)[/tex]:
[tex]\[ f(4) = \sqrt{2 \cdot 4} = \sqrt{8} \approx 2.828 \][/tex]
[tex]\[ g(4) = \sqrt{32 \cdot 4} = \sqrt{128} \approx 11.314 \][/tex]
[tex]\[ (f \cdot g)(4) = 2.828 \cdot 11.314 \approx 32 \][/tex]
Therefore, for [tex]\(x = 4\)[/tex], [tex]\((f \cdot g)(x) \approx 32.000\)[/tex].
The final result matches the calculated result for [tex]\(x = 4\)[/tex], verifying our derived function:
[tex]\[ (f \cdot g)(x) = 8x \][/tex]
1. Expression for [tex]\((f \cdot g)(x)\)[/tex]:
The notation [tex]\((f \cdot g)(x)\)[/tex] represents the product of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]. So, we need to multiply the expressions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f \cdot g)(x) = f(x) \cdot g(x) \][/tex]
2. Substitute the expressions for [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
Recall that [tex]\(f(x) = \sqrt{2x}\)[/tex] and [tex]\(g(x) = \sqrt{32x}\)[/tex]. Substituting these into the product:
[tex]\[ (f \cdot g)(x) = \sqrt{2x} \cdot \sqrt{32x} \][/tex]
3. Simplify the expression:
Using the property of square roots [tex]\(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\)[/tex], we combine the square roots:
[tex]\[ (f \cdot g)(x) = \sqrt{2x \cdot 32x} \][/tex]
4. Multiply the terms inside the square root:
Simplify [tex]\(2x \cdot 32x\)[/tex]:
[tex]\[ 2x \cdot 32x = 64x^2 \][/tex]
5. Take the square root of the resulting expression:
[tex]\[ \sqrt{64x^2} = \sqrt{64} \cdot \sqrt{x^2} = 8 \cdot x \][/tex]
Thus, the product function [tex]\((f \cdot g)(x)\)[/tex] simplifies to:
[tex]\[ (f \cdot g)(x) = 8x \][/tex]
To see this in action, let's plug in a specific value of [tex]\(x\)[/tex]:
- Let [tex]\(x = 4\)[/tex]:
[tex]\[ f(4) = \sqrt{2 \cdot 4} = \sqrt{8} \approx 2.828 \][/tex]
[tex]\[ g(4) = \sqrt{32 \cdot 4} = \sqrt{128} \approx 11.314 \][/tex]
[tex]\[ (f \cdot g)(4) = 2.828 \cdot 11.314 \approx 32 \][/tex]
Therefore, for [tex]\(x = 4\)[/tex], [tex]\((f \cdot g)(x) \approx 32.000\)[/tex].
The final result matches the calculated result for [tex]\(x = 4\)[/tex], verifying our derived function:
[tex]\[ (f \cdot g)(x) = 8x \][/tex]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.