OpenStudy (anonymous):

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

7 years ago
OpenStudy (shadowfiend):

What particular issues are you having? Just understanding the concept in general?

7 years ago
OpenStudy (anonymous):

$$I have understood the concept but getting the ans and what concept \it shows is really different 1(find the extreme values of f(x)=x^{3}-3x ^{2}-45x+25 got the solution Max=106 at x=-3 &Min value =-150 at x=5 Is \it \right ?$$

7 years ago
OpenStudy (shadowfiend):

So 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*.

7 years ago