Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To solve for [tex]\((h \circ f)(-2)\)[/tex], follow these step-by-step instructions:
1. Evaluate [tex]\( f(x) \)[/tex] for [tex]\( x = -2 \)[/tex]:
The function [tex]\( f(x) = x^3 + 7x \)[/tex].
- Substitute [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = (-2)^3 + 7(-2) = -8 - 14 = -22 \][/tex]
This gives:
[tex]\[ f(-2) = -22 \][/tex]
2. Next, evaluate [tex]\( h(x) \)[/tex] at the value found from [tex]\( f(-2) \)[/tex]:
The function [tex]\( h(x) = 6x + 1 \)[/tex].
- Substitute [tex]\( x = -22 \)[/tex]:
[tex]\[ h(-22) = 6(-22) + 1 = -132 + 1 = -131 \][/tex]
This gives:
[tex]\[ h(f(-2)) = h(-22) = -131 \][/tex]
3. Therefore, [tex]\((h \circ f)(-2) = h(f(-2)) = -131\)[/tex].
To find [tex]\( \sqrt{h(f(-2))} \)[/tex]:
- Substitute [tex]\( h(f(-2)) = -131 \)[/tex]:
[tex]\[ \sqrt{-131} \][/tex]
Since the expression inside the square root is negative, [tex]\(\sqrt{-131}\)[/tex] is not a real number. It is a complex number, specifically:
[tex]\[ \sqrt{-131} = i \sqrt{131} \][/tex]
where [tex]\( i \)[/tex] is the imaginary unit.
So, the final answer is:
[tex]\[ (h \circ f)(-2) = -131 \][/tex]
[tex]\(\sqrt{(h \circ f)(-2)} = \sqrt{-131} = i \sqrt{131} \)[/tex]
1. Evaluate [tex]\( f(x) \)[/tex] for [tex]\( x = -2 \)[/tex]:
The function [tex]\( f(x) = x^3 + 7x \)[/tex].
- Substitute [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = (-2)^3 + 7(-2) = -8 - 14 = -22 \][/tex]
This gives:
[tex]\[ f(-2) = -22 \][/tex]
2. Next, evaluate [tex]\( h(x) \)[/tex] at the value found from [tex]\( f(-2) \)[/tex]:
The function [tex]\( h(x) = 6x + 1 \)[/tex].
- Substitute [tex]\( x = -22 \)[/tex]:
[tex]\[ h(-22) = 6(-22) + 1 = -132 + 1 = -131 \][/tex]
This gives:
[tex]\[ h(f(-2)) = h(-22) = -131 \][/tex]
3. Therefore, [tex]\((h \circ f)(-2) = h(f(-2)) = -131\)[/tex].
To find [tex]\( \sqrt{h(f(-2))} \)[/tex]:
- Substitute [tex]\( h(f(-2)) = -131 \)[/tex]:
[tex]\[ \sqrt{-131} \][/tex]
Since the expression inside the square root is negative, [tex]\(\sqrt{-131}\)[/tex] is not a real number. It is a complex number, specifically:
[tex]\[ \sqrt{-131} = i \sqrt{131} \][/tex]
where [tex]\( i \)[/tex] is the imaginary unit.
So, the final answer is:
[tex]\[ (h \circ f)(-2) = -131 \][/tex]
[tex]\(\sqrt{(h \circ f)(-2)} = \sqrt{-131} = i \sqrt{131} \)[/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.