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anonymous
 5 years ago
I am having difficulty on Maxima and minima can some one explain me with an very easy eg as i am not good in my studies being frank
anonymous
 5 years ago
I am having difficulty on Maxima and minima can some one explain me with an very easy eg as i am not good in my studies being frank

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shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0What particular issues are you having? Just understanding the concept in general?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0$$I have understood the concept but getting the ans and what concept \it shows is really different 1(find the extreme values of f(x)=x^{3}3x ^{2}45x+25 got the solution Max=106 at x=3 &Min value =150 at x=5 Is \it \right ?$$

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0So the problem is: $$ f(x)=x^{3}3x^{2}45x+25 $$ And you got: Maximum = 106 @ x = 3 Minimum = 150 @ x = 5 With a quick spot check, that looks right. You can spot check first by verifying that x = 3 and x = 5 are inflection points for that function. You can determine that by finding the derivative of \(f(x)\) = \(f'(x)\) using the basic power rule. $$ \frac{d}{dx}x^3  3x^2  45x + 25 = f'(x) = 3x^2  6x  45 $$ You can plug in 3: $$ 3 \cdot 3^2  6 \cdot 3  45 = 27 + 18  45 = 0 $$ And 5: $$ 3 \cdot 5^2  6 \cdot 5  45 = 75  30  45 = 0 $$ Since the derivative is 0 at both, these two are inflection points. Once you determine that the values of \(f(x)\) at x = 2 and x = 4 are lower than at x = 3, you know that x = 3 is the *maximum*. Once you determine that the values of \(f(x)\) at x = 4 and x = 6 are higher than at = 5, you know that x = 5 is the *minimum*.
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