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Mathematics 18 Online
rvc (rvc):

find values of x,for which the function f(x)=x^3+12x^2+36x+6 is increasing

OpenStudy (cwrw238):

first find the derivative the required values are when the derivative is greater than zero

rvc (rvc):

3x^2+24x+36 = y1

rvc (rvc):

@ikram002p

OpenStudy (cwrw238):

right now equate this to zero and solve for x to find the turning points on the graph and then we can find values for x the required values of x 3(x^2 + 8x + 12) = 0

OpenStudy (cwrw238):

the function will factor

OpenStudy (cwrw238):

can you factor it?

rvc (rvc):

i did not understand

rvc (rvc):

@cwrw238

OpenStudy (cwrw238):

to factor it you need 2 numbers whose product is 12 and when added they give 8 if we call them a and b we then get (x + a)(x + b) = 0

rvc (rvc):

3(x+2)(x+6)

OpenStudy (cwrw238):

right so 3(x + 2)(x + 6) = 0 so x+2 = 0 and x +6 = 0 and x = -2 or -6

rvc (rvc):

after that

OpenStudy (cwrw238):

these are the values of x for which the graph of the function has a turning point - now we need to find the nature of these points - maximum, minimum or point of inflection

OpenStudy (cwrw238):

to do this we can find second derivative this is derivative of 3x^2 + 24x + 36 can you do that?

rvc (rvc):

6x+24

rvc (rvc):

this is y2 ,right?

OpenStudy (ikram002p):

i have to go @rational @praxer might help

rvc (rvc):

thank u green mam @ikram002p

OpenStudy (praxer):

algorithm. 1. Differentiate the function. 2. equate it to zero 3. those regions where the derivative it > 0 is the region where the function is increasing. 4. You can do so, by the wavy curve method.

OpenStudy (cwrw238):

yes now the sin of this second derivative tells you if the points are maximum of minimum x = -2: y2 = 6(-2) + 24 = 12 which id positive - so minimum x = -6 y2 = 6(-6) + 24 = -12 - maximum

rvc (rvc):

if it is +ve then mimimum? @cwrw238

OpenStudy (cwrw238):

so when x < -6 the function is increasing and also its increasing when x > -2

OpenStudy (cwrw238):

yes

rvc (rvc):

how @cwrw238

OpenStudy (praxer):

$$ f(x)= x^3+12x^2+36x+6$$ $$f'(x)=3x^2+24x+36$$ $$f'(x)=(x+6)(x+2)=0$$ The 2 and 6 is the points where the slope is 0 or the function is parallel to the x axis. now Use the wavy curve method to find the region where the function is > 0 .

OpenStudy (cwrw238):

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