Suppose the function fx has the value 3 at x2 and 2 c1 So fx

Suppose the function f(x) has the value 3 at x=2 and 2<= f \'(x) <=4 for 2<=x<=5. Use the racetrack principle to give(best possible from the information available) upper and lower bounds for f(5)

Please show work / explain the solutions

f\'(x) represents the rate of increase of f(x) So to find the max we assume f\'(x)=4. implies y=f(x)=4x+c f(2)=3 => c=-5 So f(x)=4x-3 So max(f(5))=4(5)-3=17 To find min we take f\'(x)=2. implies y=f(x)=2x+c f(2)=3 => c=-1 So f(x)=2x-1 So min(f(5))=2(5)-1=9 Hope you got it :)

$Suppose the function f(x) has the value 3 at x=2 and 2<= f \'(x) <=4 for 2<=x<=5. Use the racetrack principle to give(best possible from the informa$

Suppose the function fx has the value 3 at x2 and 2 c1 So fx

Solution

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