Answer:
[tex]f(1)=2[/tex]
Step-by-step explanation:
Given:
[tex]f(x)=\sqrt{10-x}-x[/tex]
[tex]f(x) =2[/tex]
[tex]\sqrt{10-x}-x=2\\\sqrt{10-x}=x+2[/tex]
Squaring both sides, we get
[tex](\sqrt{10-x})^{2}=(x+2)^{2}\\10-x=x^{2}+4+4x\\x^{2}+4x+x+4-10=0\\x^{2}+5x-6=0\\(x-1)(x+6)=0\\x=1\textrm{ or }x=-6[/tex]
Therefore, the possible values of [tex]x[/tex] for which [tex]f(x)=2[/tex] are -6 and 1.
Therefore, [tex]f(1)=2[/tex]
