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 February 18th, 2019, 08:48 PM #1 Member   Joined: Feb 2018 From: Iran Posts: 52 Thanks: 3 Graph of a function The graph of the function f(x) = a(x– 2)^2(x+ b) is shown in the figure.Find a and b. I solved this question by plugging f(2)=0 then I get b=-2 Then I plugged f(0)=3 and i found a but answer is not true according to my book why is this so??
 February 18th, 2019, 09:10 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 $f(2)=0$ regardless of what the value of $b$ is. Do you see why? How about if we note that $f(0)=3$ $4ab = 3$ They then seem to be trying to indicate that the maximum of $f(x)$ occurs at $x=3$ If this is the case then the first derivative of $f(x)$ will be equal to $0$ at $x=3$ $\dfrac{df}{dx} = 2a(x-2)(x+b) + a(x-2)^2 = (x-2)(2a(x+b)+a(x-2))$ $2a(x+b)+a(x-2) = 0 ~@x=3$ $2a(3+b) + a(3-2) = 0$ $6a+2ab+a = 0$ $a(7+2b)=0$ $a=0$ or $b = -\dfrac{7}{2}$ $a \neq 0$ as $4ab=3$ from above so $b = -\dfrac{7}{2}$ $4 a \left(-\dfrac{7}{2}\right) = 3$ $-14a = 3$ $a = -\dfrac{3}{14}$
February 18th, 2019, 10:13 PM   #3
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Quote:
 Originally Posted by romsek They then seem to be trying to indicate that the maximum of $f(x)$ occurs at $x=3$
This should read they seem to be trying to indicate that there is a local maximum of $f(x)$ at $x=3$

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