Differentiate (1 Viewer)

[simDS]

hi,
hey can sombody please help me.
i can diffentiate it but how do you test the answer?

'find all values of x for which the curve f(x) = x^3 - 3x + 4 is decreasing'

thanks

 

[lyounamu]

hi,
hey can sombody please help me.
i can diffentiate it but how do you test the answer?

'find all values of x for which the curve f(x) = x^3 - 3x + 4 is decreasing'

thanks

y = x^3 - 3x + 4

dy/dx = 3x^2 - 3 = 3(x^2-1) = 3(x+1)(x-1)

Draw the graph of 3(x+1)(x-1) then you will see that

3(x+1)(x-1) decreases when -1 < x < 1

NOTE: function is decreasing when dy/dx < 0 and function is increasing when dy/dx > 0

 

[munch0r]

f(x) = x^3 - 3x +4
f'(x) = 3(x^2 - 1)

if the curve is decreasing, f'(x) is negative
f'(x) is only negative when -1 < x < 1

lol beaten -__-

 

[lyounamu]

f(x) = x^3 - 3x +4
f'(x) = 3(x^2 - 1)

if the curve is decreasing, f'(x) is negative
f'(x) is only negative when -1 < x < 1

lol beaten -__-

Hehehe...no body can beat me in terms of my typing speed...:wave:

 

[Timothy.Siu]

hi,

i can diffentiate it but how do you test the answer?

u integrate it? lol

 

[simDS]

but how did you get -1 < x< 1

 

[zahmad]

but how did you get -1 < x< 1

Intercept wit x-axis

 

[simDS]

Intercept wit x-axis

so you have to draw the graph?

 

[Timothy.Siu]

but how did you get -1 < x< 1

well if u draw it, u can see clearly when its under the x-axis. but u dont hav to draw it once u find the intercepts wich are 1 and -1 u can sub in a value eg 0 so u can find if its negative or positive then u know that its less than 0 for -1<x<1