Question

the function ( f(x) ) is continuous on ( (-infty,infty) ). use the given information to sketch the graph of ( f ).

( f(0)=4,f(2)=0,f(4)= - 4 )	( f(0)=0,f(4)=0 )
( f(2)=0 )
( f(x)>0 ) on ( (2,infty) )	( f(x)<0 ) on ( (-infty,2) )

choose the correct graph of ( f ) below.
○ a.
○ b.
○ c.
○ d.

Explanation:

Step1: Analyze increasing and decreasing intervals

Since \(f^{\prime}(x)>0\) on \((-\infty,0)\) and \((4,\infty)\), the function \(f(x)\) is increasing on \((-\infty,0)\) and \((4,\infty)\). Since \(f^{\prime}(x)<0\) on \((0,4)\), the function \(f(x)\) is decreasing on \((0,4)\). So \(x = 0\) is a local maximum (because the function changes from increasing to decreasing at \(x = 0\)) and \(x=4\) is a local minimum (because the function changes from decreasing to increasing at \(x = 4\)).

Step2: Analyze concavity

Since \(f^{\prime\prime}(x)<0\) on \((-\infty,2)\), the function \(f(x)\) is concave - down on \((-\infty,2)\). Since \(f^{\prime\prime}(x)>0\) on \((2,\infty)\), the function \(f(x)\) is concave - up on \((2,\infty)\). And \(x = 2\) is an inflection point (\(f^{\prime\prime}(2)=0\) and concavity changes at \(x = 2\)).

Step3: Use function values

We know that \(f(0)=4\), \(f(2)=0\), \(f(4)=-4\)

Now let's check each option:

• Option A:
• At \(x = 0\), it is a local maximum (\(y = 4\)), at \(x=4\) it is a local minimum (\(y=-4\)). The function is increasing on \((-\infty,0)\) and \((4,\infty)\), decreasing on \((0,4)\). It is concave - down on \((-\infty,2)\) and concave - up on \((2,\infty)\)
• Option B: The function is decreasing on \((-\infty,0)\) which is wrong as \(f^{\prime}(x)>0\) on \((-\infty,0)\)
• Option C: The function is increasing on \((0,4)\) which is wrong as \(f^{\prime}(x)<0\) on \((0,4)\)
• Option D: The function is decreasing on \((4,\infty)\) which is wrong as \(f^{\prime}(x)>0\) on \((4,\infty)\)

Answer:

A.