The first step is to chop up the expression into two terms: . Now, integrate each term. Whenever there is an x on the denominator, the integral is lnx. Multiply that by the numerator of 4. The integral of 1 is x. Put those together to get your answer of .

To find where the local minima occur for the function, we need the x-values at which the first derivative changes from negative to positive. 

The first derivative is

and was found using the following rule:

Now, we must find the critical values at which the first derivative is equal to zero:

Now, using the critical values, we make the intervals on which we see whether the first derivative is positive or negative (plug in any value on the interval into the first derivative function):

On the first interval, the first derivative is positive, on the second it is negative, and on the third it is positive. So, a local minimum occurs at

because the first derivative changed from positive to negative at that point.

We need to differentiate term by term, applying the power rule,

This gives us

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

Any local minimum will fall at a critical point where the derivative passes from negative to positive. To check this, we check a point in each of the intervals defined by the critical points:

.

Let's take -2 from the first interval, 0 from the second interval, and 2 from the third interval.

The derivative moves from negative to positive at 1, so that is the function's only local minimum.

To determine the local minima for the function, we must determine where the first derivative changes from negative to positive.

The first derivative of the function is equal to

and was found using the following rules:

, 

Now, we must find the critical values, the values at which the first derivative equals zero:

We use the critical value to make our intervals on which we check the sign of the first derivative:

Note that at the endpoints of the intervals, the first derivative is neither positive nor negative.

Next, we plug in any value on the intervals into the first derivative and check the sign. On the first interval, the first derivative is negative, while on the second, it is positive. Thus, a local mininum exists at .

First, take the derivative of the function, remembering to multiply the exponent by the coefficeint and then subtracting one from the exponent. Therefore, the derivative is: . Then, set it equal to 0 to get your critical point: . Then, test a value on either side of your critical point and plug into the derivative. To the left, the value is negative. To the right, the value is positive. Therefore, your local minimum is at .

To find the local minimum of a function, one must find the critical points (where the derivative is equal to zero).

Taking the derivative of the function above using the power rule,  

.

The derivative is 

.

Setting this equal to zero and solving for x gets you 

To see if this indeed is a local minimum one must plug in values below and above this value of x to see if the derivative is positive (increasing) or negative (decreasing) around that point.

If you plug -2 into the equation for  you get -18 (decreasing), and plugging in 0, you get  (increasing).

Thus, is a local minimum.

This function has only three extrema: a local maximum at  and two minima at  (these extrema are found by finding the first derivative of the function, setting it equal to zero, and solving for x).

By evaluating any point along the four intervals defined by the extrema, , , , and , one can see that the function is decreasing on the first interval, increasing on the second, decreasing on the third and increasing on the fourth.  Therefore,  is a local maximum while  are minima Furthermore, the value of the function is the same at both minima, making them global minima.

The clearest indicators to check these answers with are (1) points where a function crosses zero compared to where another function has zero slope, and (2) where a function is increasing compared to where another function is positive.

Using the first criteria, we can see that f(x) has a slope of zero where both g(x) and h(x) cross zero, meaning f'(x) is not i(x).  

Using the second criteria, we can see that f(x) is increasing on the entire interval shown, while g(x) is positive on the entire interval.  Therefore, f'(x)=g(x).

The condition reads 'the limit of f of x as x approaches three from the left'.  Note that we're not looking at x as it approached negative three.  Only A and B have infinite limites at positive 3 (C and D show limits as x approaches negative 3), so the answer must be one of these.  We can see that from the left, B approaches positive infinity at x=3 from the left, while A approaches negative infinity at x=3 from the left, so the correct answer is A.

Since this is a quadratic function with a positive first coefficient, we know it only has one local minimum and no other extrema. So all we have to do is find the x coordinate corresponding to this value and we will have one endpoint of our interval, and it will be decreasing constantly as you approach this point from the left.

Recall, 

So, if we take the derivative of this quadratic, we get

So this is where our interval ends. Since there are no other local extrema, we know our interval is 

.