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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]
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