Suppose $$ f $$ is continuous on $$ [1,5] $$ and the only solutions of the equation $$ f(x)= $$ 6 are $$ x=1 $$ and $$ x=4 $$. If $$ f(2)=8 $$, explain why $$ f(3)6 $$. Show that $$ F(x)=|x| $$ is continuous evervwhere.

Question

Suppose $$ f $$ is continuous on $$ [1,5] $$ and the only solutions of the equation $$ f(x)= $$ 6 are $$ x=1 $$ and $$ x=4 $$. If $$ f(2)=8 $$, explain why $$ f(3)>6 $$. Show that $$ F(x)=|x| $$ is continuous evervwhere.

Lambert Brooks · Accepted Answer

$$ f(3) > 6 $$ because $$ f $$ is continuous and $$ f(2) = 8 $$, which is greater than 6. $$ F(x) = |x| $$ is continuous everywhere because it is continuous for all real numbers.

Suppose \( f \) is continuous on \( [1,5] \) and the only solutions of the equation \( f(x)= \) 6 are \( x=1 \) and \( x=4 \). If \( f(2)=8 \), explain why \( f(3)>6 \). Show that \( F(x)=|x| \) is continuous evervwhere.

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Suppose \( f \) is continuous on \( [1,5] \) and the only solutions of the equation \( f(x)= \) 6 are \( x=1 \) and \( x=4 \). If \( f(2)=8 \), explain why \( f(3)>6 \). Show that \( F(x)=|x| \) is continuous evervwhere.

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Solución paso a paso

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