Question

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

Answer 1

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

Answer 2

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

Answer 3

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*.