Question

question
the graph of the function ( f ) is shown above. if ( a ), ( b ), and ( c ) are values of ( x ) for which ( f(x) ) is defined, which following must be true?
( \bigcirc f(0)+f(a)=f(a) )
( \bigcirc ) if ( a < b ), then ( f(a)<f(b) ).
( \bigcirc ) if ( a < 0 ), then ( f(a)<0 ).
( \bigcirc ) if ( a + b = c ), then ( f(a)+f(b)=f(c) ).

Explanation:

Step1: Analyze \(f(0)+f(a)=f(a)\)

From the graph, when \(x = 0\), \(y=f(0)=0\). Then \(f(0)+f(a)=0 + f(a)=f(a)\)

Step2: Analyze "If \(a < b\), then \(f(a)<f(b)\)"

The function is not strictly increasing for all \(x\). For example, if \(a\) is in the negative - \(x\) region and \(b\) is in the positive - \(x\) region, the relationship \(f(a)<f(b)\) does not hold for all \(a < b\) (since the function has two different linear - like parts with different slopes)

Step3: Analyze "If \(a < 0\), then \(f(a)<0\)"

From the graph, when \(x<0\), \(y = f(x)>0\). So the statement "If \(a < 0\), then \(f(a)<0\)" is false

Step4: Analyze "If \(a + b=c\), then \(f(a)+f(b)=f(c)\)"

Let \(f(x)=mx\) for \(x\geq0\) and \(f(x)=nx\) for \(x < 0\) (\(m
eq n\)). If \(a>0\), \(b>0\), \(c=a + b\), then \(f(a)+f(b)=ma+mb=m(a + b)=f(c)\). But if \(a<0\) and \(b>0\) (or vice - versa), \(f(a)+f(b)=na+mb\) and \(f(c)=f(a + b)\). Since \(n
eq m\), \(na+mb
eq f(a + b)\) in general

Answer:

\(f(0)+f(a)=f(a)\)