For the given function
[tex] f(x) = -4 -x + 3x^2 [/tex]-----------------------(1)
we have to see when it will increase and decrease.
For that we will first find derivative of function
Now derivative of first term in f(x) which is -4 will be 0 as its constant.
then derivative of x is 1, so derivative of -x will be -1
To derivate last term of function, [tex] 3x^{2} [/tex], we will use power rule formula:
[tex] (x^{n} )'= nx^{n-1} [/tex]
so [tex] (x^{2} )'= 2x^{2-1} = 2x^{1} = 2x [/tex]
constant 3 will come as it is, so derivative of 3x^2 wil be 3(2x)= 6x.
So f'(x) = 0 -1 +6x
f'(x) = -1 + 6x------------------------------(2)
For function to be increasing f'(x) should be positive and for function to be decreasing f'(x) shoould be negative
So we will first find where f'(x) = 0
So put 0 in f'(x) place in equation (2)
0 = -1 + 6x
Now solve for x as shown
0 +1 = -1 + 6x +1
1 = 6x
[tex] \frac{1}{6} = \frac{6x}{6} [/tex]
[tex] \frac{1}{6} = x [/tex]
So now we can have two regions as shown on either side of 1/6 on number line
_____________I______________[tex] \frac{1}{6} [/tex]_______II__________
So to test region I, pick any number to the left side of [tex] \frac{1}{6} [/tex]. For example lets take 0. now plug 0 inx place in f'(x) equation given by (2)
f'(x)= -1 + 6x
f'(0) = -1 + 6(0) = -1+0 = -1 which is negative. So since f'(x) we got as negative for region I, so this will be decreasing.
Interval for region I will be (-∞, [tex] \frac{1}{6} [/tex])
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similarly now test region II. For that pick any number to the right of [tex] \frac{1}{6} [/tex], lets take 1. So plug 1 in x place in f'(x) equation given by (2)
f'(x) = -1 + 6x
f'(1) = -1 +6(1)= -1+6 = 5 which is positive so we will have f(x) increasing in this region.
Interval for region II will be ([tex] \frac{1}{6} [/tex], ∞)
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Graphs of [tex] f(x) = -4 -x + 3x^2 [/tex] and [tex] f'(x) = -1 +6x [/tex]are shown in attachment. So you can clearly see in graph that f'(x) is negative (below x axis) from -∞ till [tex] \frac{1}{6} [/tex] so f(x) should be decreasing in this part which we can see from f(x) graph that its decreasing from -∞ till [tex] \frac{1}{6} [/tex].
Similarly f'(x) is positive( above x axis) beyond [tex] \frac{1}{6} [/tex] so its increasing beyond [tex] \frac{1}{6} [/tex] which we can veryify from f(x) graph we can see that its increasing from [tex] \frac{1}{6} [/tex] till ∞. Hence verified
