To find the equation for the tangent line, we must first find the derivative of \(\displaystyle f(x)\).

In order to evaluate this derivative, we must use these formulae:

\(\displaystyle \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1, \frac{d}{dx}cx=c\)

\(\displaystyle \frac{d}{dx}(x^{n}))=nx^{n-1}, \frac{d}{dx}(cx^{n})=ncx^{n-1}\)

\(\displaystyle f'(x)=3*3x^{3-1}-2x^{2-1}+0=9x^{2}-2x\)

Next we find the slope of the tangent line at this point:

\(\displaystyle f'(-2)=9(-2)^{2}-2(-2)=36+4=40\)

Then we need to plug the given \(\displaystyle x\) value back into the original function in order to find the corresponding \(\displaystyle y\) value.

\(\displaystyle f(-2)=3*(-2)^{3}-(-2)^{2}+1=-24-4+1=-27\)

Now that we have a point \(\displaystyle (-2,-27)\) and a slope of \(\displaystyle 40\), we can find the equation for the line by plugging these values into  \(\displaystyle y=mx+b\)  form to solve for \(\displaystyle b\):

\(\displaystyle -27=40*-2+b\) ==> \(\displaystyle b = 53\)

So, the equation for the line is \(\displaystyle y=40x+53\).

For this equation, let's start with acceleration and move backwards towards position. We know that the object is accelerating at a constant rate of \(\displaystyle -9.8\)

\(\displaystyle p''(t)=-9.8\)

To solve for velocity , we integrate both sides with respect to \(\displaystyle t\)

\(\displaystyle \int p''(t)dt=\int -9.8 dt\)

\(\displaystyle p'(t)=-9.8t+B\), where \(\displaystyle B\) is a constant

Since we know that the initial velocity is \(\displaystyle 13\), \(\displaystyle p'(0)=13, B=13\). 

\(\displaystyle p'(t)=-9.8t+13\)

To solve for position we have to take the integral of both sides with respect to \(\displaystyle t\).

\(\displaystyle \int p'(t)dt=\int (-9.8t+13)dt\)

\(\displaystyle p(t)=-4.9t^2+13t+C\), where \(\displaystyle C\) is a constant. 

Since we know that the initial position is \(\displaystyle 13\), \(\displaystyle P(0)=35\), \(\displaystyle C=35\)

The formula for this motion can be modelled by:

\(\displaystyle p(t)=-4.9t^2+13t+35\)

\(\displaystyle 4y-x=4\) can be written as \(\displaystyle y=\frac{1}{4}x+1\)with slope of \(\displaystyle \frac{1}{4}\) and y-intercept of 1.  So the inverse function would be \(\displaystyle x=\frac{1}{4}y+1\)or, \(\displaystyle x-{\frac14}y=1\)

\(\displaystyle \ln x=-9\) implies \(\displaystyle e^{lnx}=e^{-9}\) implies \(\displaystyle x=e^{-9}\)

To write the expression, simply use the bounds laid out by the x-coordinates of the points given. Thus, your answer is:

\(\displaystyle \int_{-3}^{5}(4x^3-2x+7)dx\)

Both \(\displaystyle cx^2-2x\) and \(\displaystyle x\) are continuous in the intervals they are defined. In order for the function to be continuous on the entire real line, we need to have \(\displaystyle cx^2 -2x =x\) at \(\displaystyle x =3\). Plugging the value in, we get \(\displaystyle c \cdot 3^2 - 2 \cdot 3 = 3\) \(\displaystyle \Rightarrow 9c-6 =3 \Rightarrow c=1.\)

To solve, simply differentiate and don't forget to indicate the differentiation by the "apostrophe" between the function and dependent variable. Thus,

\(\displaystyle p'(t)=4t\)

To solve, simply differentiate \(\displaystyle p(t)\).

Remember to use the power rule.

Recall the power rule:

\(\displaystyle f(x)=x^n\)

\(\displaystyle f'(x)=nx^{n-1}\)

Apply this to our situation to get

\(\displaystyle v(t)=p'(t)=10t-6\)

The second derivative is the derivative of the first derivative function.

With our original function, \(\displaystyle g(x)=e^{2x}-Sin(x)+40\)

We may find the first derivative to be: \(\displaystyle g'(x)=2e^{2x}-Cos(x)\)

Note that the chain rule was used to find the first derivative.

Now to find the second derivative we must take the derivative of this function:

\(\displaystyle g''(x)=4e^{2x}+Sin(x)\)

To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 3. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 100. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.