In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= 2(x+2)^2 + 3x\)

That is done by doing the following: \(\displaystyle f'(x)= 4x+11\)

Then plug \(\displaystyle x=3\) into the derivative: \(\displaystyle f'(3)= 4({\color{Blue} 3})+11\)

Therefore, the answer is: \(\displaystyle 23\)

We notice that this function,

\(\displaystyle f(x)=e^xSin(x)\), is of the form \(\displaystyle f(x)=g(x)h(x)\) 

such that we must use the Product Rule to find the derivative.

\(\displaystyle f'(x)=g(x)h'(x)+h(x)g'(x)\).

Doing so we find the derivative to be:

\(\displaystyle f'(x)=e^xCos(x)+e^xSin(x)=e^x(Cos(x)+Sin(x))\)

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle A=4\pi r^2\)

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}\)

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of shrinkage of the volume is 18.5 times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=(18.5)8\pi r \frac{dr}{dt}\)

\(\displaystyle r =(18.5)2\)

\(\displaystyle r=37\)

The diameter is then:

\(\displaystyle d=2r\)

\(\displaystyle d=74\)

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle A=4\pi r^2\)

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}\)

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of shrinkage of the volume is \(\displaystyle \frac{3}{\pi}\) times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=(\frac{3}{\pi})8\pi r \frac{dr}{dt}\)

\(\displaystyle r =(\frac{3}{\pi})2\)

\(\displaystyle r=\frac{6}{\pi}\)

The circumference of a sphere is:

\(\displaystyle C=2\pi r\)

\(\displaystyle C=2\pi(\frac{6}{\pi})\)

\(\displaystyle C=12\)

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle A=4\pi r^2\)

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}\)

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 0.16 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=(0.16)8\pi r \frac{dr}{dt}\)

\(\displaystyle r =(0.16)2\)

\(\displaystyle r=0.32\)

Finding the volume:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle V=\frac{4}{3}\pi (0.32)^3=0.137\)

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle A=4\pi r^2\)

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}\)

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 12.5 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=(12.5)8\pi r \frac{dr}{dt}\)

\(\displaystyle r =(12.5)2\)

\(\displaystyle r=25\)

Begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle C=2\pi r^\)

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}\)

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now given the problem information, the rate of growth of the volume is 98 times the rate of growth of the circumference, solve for the radius:

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=(98)2\pi \frac{dr}{dt}\)

\(\displaystyle r^2=(98)\frac{1}{2}\)

\(\displaystyle r =\sqrt{\frac{98}{2}}=\sqrt{49}=7\)

Begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle C=2\pi r^\)

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}\)

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now given the problem information, the rate of growth of the volume is 72 times the rate of growth of the circumference, solve for the radius:

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=(72)2\pi \frac{dr}{dt}\)

\(\displaystyle r^2=(72)\frac{1}{2}\)

\(\displaystyle r =\sqrt{\frac{72}{2}}=\sqrt{36}=6\)

Finding the volume at this time:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle V=\frac{4}{3}\pi (6)^3=288\pi\)

Begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle C=2\pi r^\)

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}\)

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now given the problem information, the rate of shrinkage of the volume is 338 times the rate of shrinkage of the circumference, solve for the radius:

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=(338)2\pi \frac{dr}{dt}\)

\(\displaystyle r^2=(338)\frac{1}{2}\)

\(\displaystyle r =\sqrt{\frac{338}{2}}=\sqrt{169}=13\)

The circumference is then:

\(\displaystyle C=2\pi r\)

\(\displaystyle C=2\pi (13)=26\pi\)

Begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

\(\displaystyle C=2\pi r^\)

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}\)

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now given the problem information, the rate of shrinkage of the volume is 450 times the rate of shrinkage of the circumference, solve for the radius:

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=(450)2\pi \frac{dr}{dt}\)

\(\displaystyle r^2=(450)\frac{1}{2}\)

\(\displaystyle r =\sqrt{\frac{450}{2}}=\sqrt{225}=15\)

The diameter is then:

\(\displaystyle d=2r\)

\(\displaystyle d=30\)