how do you find the local maximun or minimun of a fuction without using a graphing calculator?

Question

for example:

f(x) = x^3 + 2x^2 - 5x + 1

there is one local maximun and one local minimun. How do you find them without using a graphing calculator?

Answer 1

(1) Take the first derivative, set it equal to zero and find the zeroes. Those are local extrema.
(2) Take the second derivative and plug in the points you found in (1). If you get a negative number (meaning the original function is curved concave down), it's a local maximum. If you get a positive number (meaning the original function is curved concave up), it's a local minimum.

1) f(x) = x^3 + 2x^2 -5x +1
f ' (x) = 3x^2 + 4x -5
0 = 3x^2 + 4x -5
x= [-2+/- sqrt(19)] / 3

2) f '' (x) = 6x + 4
plugging in [-2+sqrt(19)]/3 will give a positive number because sqrt(19) is bigger than 2.

plugging in [-2-sqrt(19)]/3 will give a negative number because that quantity is negative and multiplied by 6 is bigger than 4.

So there is a local minimum at [ -2 + sqrt(19)] / 3
and there is a local maximum at [ -2 - sqrt(19)] / 3

Answer 2

Take the derivative and set it to 0, getting:
3x^2 +4x -5 = 0
x = [-4 +/- sqrt(16+60)]/(2*3)
x = [-4 +/- 2sqrt(19)]/6
x = - 2/3 +/- 1/3sqrt(19)This gives you the two places where
f(x) has a horizontal tangent and hence a max or min. Now you can use the second derivative to determine which it is, or you can just take a value on euther side of the root to see iwhat happens to y as you pass from one side of x to the other.

Answer 3

start by taking the derivative, when you find the derivative and it is correct, equal it to zero and solve, those will be points on the graph where the slope is a zero or 180 degrees.

1) now that they are solved, setup a interval line, determine if it is negative from a negative infinity to the number you found when solved; also do the same for the number you found out to infinity. kind of like taking a limit to the number you found and seeing whether or not it is positive or negative.
2) when you find out if the graph comes in from negative infinity is negative you have a local or extrema maxi ma (if open interval) if it is positive it will be a maximum.

Answer 4

take the derivative and set it equal to zero the solve the resulting quadratic.

3x^2 + 4x - 5 = 0

Answer 5

Take the first derivative and set equal to zero and solve for x.
.

Answer 6

get the first derivative

f'(x)=3x^2+4x-5

put f'(x)=0

3x^2+4x-5=0
find the points , substitute in f''(x) , if -ve , local max , if +ve , local minimum


hope i helped

Answer 7

You need calculus. The way they're found is using the first or second derivative test.

Answer 8

By solving equation

f'(x) = 0
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