" is decreasing at ." is incorrect.  The function is increasing at because .

" is increasing over the interval ." is possibly true, but there is not enough information to conclude that it must be true.

" has a point of inflection at ." is possibly true.  Although we know that , a requirement for an inflection point, we do not know that changes signs at .

" must have at least one relative maximum." is possibly true, but there is not enough information to conclude that it must be true.

"The line is tangent to ." must be true.  Because , the function travels through the point .  Because , the slope of the line tangent to the curve at is 5.  Use point-slope form to determine the equation of the tangent line.

The derivative of the function is , and is found using the power rule

and the rule for the derivative of natural log which is, 

so plugging in  gives , which must be the slope of the line since the tangent line's slope is determined by the derivative.

Thus, the line is of the form , where b is unknown.

Solve for b by setting the equation equal to  and plugging in  for x since that is the given point. 

, which gives us 

In order to find the slope of the tangent line through a certain point, we must find the rate of change (derivative) of the function.  The derivative of   is written as .  This tells us what the slope of the tangent line is through any point  in our function .  In other words, all we need to do is plug-in  (because our point  has an x-value of 1) into .  This will give us our answer,  .

To find the tangent line one must first find the slope, this can be given by the derivative evaluated at a point.

To find the derivative of this function use the power rule which states,

The derviative of  is .

Evaluated at our point ,

we find that the slope, m is also 3.

Now we may use the point-slope equation of a line to find the tangent line.

The point slope equation is 

 Where  is the point at which the line is tangent.

Using this definition we find the tangent line to be defined by .

The first derivative is easy:

f'(x) = 12x³ – 15x² – 4

The slope of the tangent line is found by calculating f'(40) = 12 * 40³ – 15 * 40² – 4 = 768,000 – 24,000 – 4 = 743,996

To find the slope of a tangent line, we need to find the first derivative of the function at that point. In other words, we need y'(6).

Taking the first derivative using the Power Rule  we get the following.

Substituting in 6 for b and solving we get:

.

So our answer is 320160

To find the slope of a tangent line, we need the first derivative.

Recall that to find the first derivative of a polynomial, we need to decrease each exponent by one and multiply by the original number.

The slope of the tangent line can be found easily via derivatives. To find the slope of the tangent line at s=16, find b'(16) using the power rule on each term which states:

Applying this rule we get:

Therefore, the slope we are looking for is 454.

To find the slope of the line at that point, find the derivative of f(x) and plug in that point. 

Remember that the derivative of  and the derivative of   

Now plug in  

The first step here is to integrate  in order to get .

Here the problem tells us that the integration constant , so

Plug in  here