Question

how to find maxima and minima of a given function with derivatives explain it with an example.( absolute extreme)
plz don't copy paste
tomorrow I m having exam
so plz dont try to cheat ​

Answer 1

Answer:

Step-by-step explanation:

The First Derivative: Maxima and Minima

Consider the function

f(x)=3x4−4x3−12x2+3

on the interval [−23]. We cannot find regions of which f is increasing or decreasing, relative maxima or minima, or the absolute maximum or minimum value of f on [−23] by inspection. Graphing by hand is tedious and imprecise. Even the use of a graphing program will only give us an approximation for the locations and values of maxima and minima. We can use the first derivative of f, however, to find all these things quickly and easily.

Increasing or Decreasing?

Let f be continuous on an interval I and differentiable on the interior of I.

If f(x)0 for all xI, then f is increasing on I.

If f(x)0 for all xI, then f is decreasing on I.

Example

The function f(x)=3x4−4x3−12x2+3 has first derivative

f(x) = = = 12x3−12x2−24x 12x(x2−x−2) 12x(x+1)(x−2)  

Thus, f(x) is increasing on (−10)(2) and decreasing on (−−1)(02).

Relative Maxima and Minima

Relative extrema of f occur at critical points of f, values x0 for which either f(x0)=0 or f(x0) is undefined.

First Derivative Test

Suppose f is continuous at a critical point x0.

If f(x)0 on an open interval extending left from x0 and f(x)0 on an open interval extending right from x0, then f has a relative maximum at x0.

If f(x)0 on an open interval extending left from x0 and f(x)0 on an open interval extending right from x0, then f has a relative minimum at x0.

If f(x) has the same sign on both an open interval extending left from x0 and an open interval extending right from x0, then f does not have a relative extremum at x0.

In summary, relative extrema occur where f(x) changes sign.

Example

Our function f(x)=3x4−4x3−12x2+3 is differentiable everywhere on [−23], with f(x)=0 for x=−102. These are the three critical points of f on [−23]. By the First Derivative Test, f has a relative maximum at x=0 and relative minima at x=−1 and x=2.

Absolute Maxima and Minima

If f has an extreme value on an open interval, then the extreme value occurs at a critical point of f.

If f has an extreme value on a closed interval, then the extreme value occurs either at a critical point or at an endpoint.

According to the Extreme Value Theorem, if a function is continuous on a closed interval, then it achieves both an absolute maximum and an absolute minimum on the interval.

Example

Since f(x)=3x4−4x3−12x2+3 is continuous on [−23], f must have an absolute maximum and an absolute minimum on [−23]. We simply need to check the value of f at the critical points x=−102 and at the endpoints x=−2 and x=3:

f(−2) f(−1) f(0) f(2) f(3) = = = = = 35 −2 3 −29 30  

Thus, on [−23], f(x) achieves a maximum value of 35 at x=−2 and a minimum value of -29 at x=2.