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Sagot :
The second function's output exceeds (get bigger than) the first function's output for x∈(-∞,[tex]\frac{1-\sqrt{29} }{2}[/tex])∪([tex]\frac{1+\sqrt{29} }{2}[/tex],∞)
Given the functions f(x)=x+7 and f(x)=[tex]x^{2}[/tex]
To know when to exceed, need to equate both functions.
⇒[tex]x^{2} =x+7[/tex]
⇒[tex]x^{2} -x-7=0[/tex]
x=[tex]\frac{-b+\sqrt{D} }{2a}[/tex] and [tex]\frac{-b-\sqrt{D} }{2a}[/tex]
⇒x=[tex]\frac{1-\sqrt{29} }{2}[/tex],[tex]\frac{1+\sqrt{29} }{2}[/tex]
f(x)=[tex]x^{2}[/tex] decreases from -∞ to 0 and increases from 0 to ∞ it cuts the graph of x+7 two times and for x∈(-∞,[tex]\frac{1-\sqrt{29} }{2}[/tex])∪([tex]\frac{1+\sqrt{29} }{2}[/tex],∞) f(x)=[tex]x^{2}[/tex] is greater than f(x)=x+7
Therefore,The second function's output exceeds (get bigger than) the first function's output for x∈(-∞,[tex]\frac{1-\sqrt{29} }{2}[/tex])∪([tex]\frac{1+\sqrt{29} }{2}[/tex],∞)
Learn more about function here:
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