Mathematics, 14.07.2019 06:30 chloeethoma24

Find the intervals on which f is increasing and decreasing. superimpose the graphs of f and f prime to verify your work. f(x)= -4 - x + 3x^2

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Answer from: keisha7795

For the given function

f(x) = -4 -x + 3x^2 -----------------------(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, $3x^{2}$ , we will use power rule formula:

$(x^{n} )'= nx^{n-1}$

so $(x^{2} )'= 2x^{2-1} = 2x^{1} = 2x$

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

$\frac{1}{6} = \frac{6x}{6}$

$\frac{1}{6} = x$

So now we can have two regions as shown on either side of 1/6 on number line

_____________I______________ $\frac{1}{6}$ _______II__________

So to test region I, pick any number to the left side of $\frac{1}{6}$ . 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 (-∞, $\frac{1}{6}$ )

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similarly now test region II. For that pick any number to the right of $\frac{1}{6}$ , 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 ( $\frac{1}{6}$ , ∞)

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Graphs of f(x) = -4 -x + 3x^2 and f'(x) = -1 +6x are shown in attachment. So you can clearly see in graph that f'(x) is negative (below x axis) from -∞ till $\frac{1}{6}$ so f(x) should be decreasing in this part which we can see from f(x) graph that its decreasing from -∞ till $\frac{1}{6}$ .

Similarly f'(x) is positive( above x axis) beyond $\frac{1}{6}$ so its increasing beyond $\frac{1}{6}$ which we can veryify from f(x) graph we can see that its increasing from $\frac{1}{6}$ till ∞. Hence verified