First, we must find the first derivative of f(x). This is simple:

f'(x) = 8x – 5

Now, the relative extreme is found at the point that the derivative is equal to 0.  Therefore, set f'(x) = 0:

0 = 8x – 5; 8x = 5; x = 5/8

Now, plug that into our original equation:

f(5/8) = 4(5/8)² – 5(5/8) + 20 = (100/64) – (25/8) + 20 = (100 – 200)/64 + 20 = (–100/64) + 20 = (1280 – 100)/64 = 1180/64 = 295/16

Take the derivative of the function with respect to \(\displaystyle x\):

\(\displaystyle f'=2x+b\)

Plug in the point \(\displaystyle (2,0)\) and solve for \(\displaystyle b\):

\(\displaystyle 2(2)+b=0\)

\(\displaystyle b=-4\)

Now the function is \(\displaystyle f(x)=x^2-4x+c\).

To find \(\displaystyle c\), plug \(\displaystyle (2,0)\) into the original function and solve for \(\displaystyle c\):

\(\displaystyle 0=2^2-4(2)+c\)

\(\displaystyle c=4\)

A local minimum occurs when \(\displaystyle y'\) changes from negative to positive. The first step is find \(\displaystyle y'\).

\(\displaystyle y'=cos(x)\)

Next, find the critical points (when \(\displaystyle y'=0\) or undefined).

\(\displaystyle x=\frac{\pi }{2},\frac{3\pi }{2}\)

The final step is to test on which intervals \(\displaystyle y'\) is negative and positive to identify at which of these \(\displaystyle x\) values the minimum occurs. Notice here that the first value of \(\displaystyle \frac{\pi}{2}\) falls outside of the region specified in the problem.

The regions to be tested are \(\displaystyle \left(\pi ,\frac{3\pi }2\right)\) and \(\displaystyle \left(\frac{3\pi }{2},2\pi \right)\).

To test the regions, choose a value in that region and plug it into \(\displaystyle y'\) to see if it is positive or negative.

\(\displaystyle y'\) is negative in the first region and positive in the second region, so the minimum occurs at \(\displaystyle \frac{3\pi }{2}\).

To find the corresponding \(\displaystyle y\) value, plug in this value into the original \(\displaystyle y\) function, giving us a value of \(\displaystyle -1\).

If \(\displaystyle \frac{dy}{dx}=0\) at some point \(\displaystyle x=a\), then \(\displaystyle a\) might be a local maxima.

\(\displaystyle \frac{dy}{dx}=6x^2+6x-36=0\)

\(\displaystyle 6(x+3)(x-2)=0\)

So the possible solutions are \(\displaystyle x=-3, x=2\)

In order to determine if these points are maxima (or minima) we need to determine if the function is concave up (minima) or concave down(maxima) at these points. To do this we need the second derivative. 

\(\displaystyle \frac{d^2y}{dx^2}>0 \rightarrow Concave\ Up\)

\(\displaystyle \frac{d^2y}{dx^2}< 0 \rightarrow Concave\ Down\)

Taking the second derivative of our function we get: 

\(\displaystyle \frac{d^2y}{dx^2}=12x+6\)

Plugging in the x values we want to examine we see the following.

\(\displaystyle \frac{d^2y}{dx^2}\mid_{x=-3}=-30 < 0\rightarrow\)maxima

\(\displaystyle \frac{d^2y}{dx^2}\mid_{x=2}=30 >0\rightarrow\)minima

Therefore the only local maxima is at the point \(\displaystyle x=-3\)

To find the local minimum of any graph, you must first take the derivative of the graph equation, set it equal to zero and solve for \(\displaystyle x\).  To take the derivative of this equation, we must use the power rule,  

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

We also must remember that the derivative of a constant is 0. Taking the derivative of the graph equation, we obtain the slope equation \(\displaystyle y^{'}=4x-8\). Solving for x when the equation is set to 0, we obtain \(\displaystyle x=2\).  

In order for \(\displaystyle x\) to be a local minimum, the slope must increase as it passes 2 from the left.  Plugging in 1 and 3 into the slope equation, we find that the slope is in fact increase from -4 to 4, therefore \(\displaystyle x=2\) is a local minimum.  

Plugging \(\displaystyle x=2\) back into the original graph equation to solve for \(\displaystyle y\), we find the coordinates of the local minimum for this graph is in fact \(\displaystyle (2,0)\).

Find any local maxima or minima of f(x) on the interval [-10,10]

\(\displaystyle f(x)=3x^2+5x-4\)

To begin finding local mins and maxes we need to take the first derivative of the above function.

\(\displaystyle f'(x)=6x+5\)

Local minimum and maximums occur wherever the first derivative is 0. 

\(\displaystyle 0=6x+5\)

\(\displaystyle x=\frac{-5}{6}\) 

Find the y coordinate via:

\(\displaystyle f\left(\frac{-5}{6}\right)=3\left(\frac{-5}{6}\right)^2+5\left(\frac{-5}{6}\right)-4=-6\frac{1}{12}\)

So the first bit of our answer:

\(\displaystyle \left(\frac{-5}{6},-6\frac{1}{12}\right)\)

But is it a maximum or minimum?

To find that, we need to know if the function is concave up or concave down at the point.

To test concavity we need the second derivative:

\(\displaystyle f''(x)=6\)

The second derivative is positive everywhere, so this function is concave up everywhere, making this a local minimum.

First, find the derivative of the function. 

\(\displaystyle y'=x(3x+2)\).

Then we find that the points where the derivative equals 0 are at \(\displaystyle \frac{-3}{2}\) and \(\displaystyle 0\). Then picking points on the intervals between these critical points and plugging them into the derivative. Where the results are negative, the function is decreasing. Where they are positive, the function in increasing.

Using this method we see that in the on the interval \(\displaystyle \left \left( -\infty, -\frac{2}{3} \right)\) the function is increasing. On \(\displaystyle \left ( \frac{-2}{3},0 \right )\) the function is decreasing. On \(\displaystyle \left ( 0,\infty \right )\) the function is increasing. Because the function is increasing to the left of \(\displaystyle -\frac{2}{3}\) and decreasing to the right, \(\displaystyle x=-\frac{2}{3}\) is a local maximum. Plugging it into the function, we can find the y value.      

First, find the derivative of the function. 

\(\displaystyle y'=x(3x+2)\).

Then we find that the points where the derivative equals 0 are at \(\displaystyle \frac{-3}{2}\) and \(\displaystyle 0\). Then picking points on the intervals between these critical points and plugging them into the derivative. Where the results are negative, the function is decreasing.

Where they are positive, the function in increasing. Using this method we see that in the on the interval \(\displaystyle \left ( -\infty , -\frac{2}{3} \right)\) the function is increasing. On \(\displaystyle \left ( \frac{-2}{3},0 \right )\) the function is decreasing. On \(\displaystyle \left ( 0,\infty \right )\) the function is increasing. Because the function is decreasing to the left of \(\displaystyle 0\) and increasing to the right, \(\displaystyle x=0\) is a local minimum. Plugging it into the function, we can find the y value.      

\(\displaystyle f(-1)=(-1)^3+4(-1)^2+5(-1)\) To find the least y-value, we must first find where the minimum of the function is. This is achieved by finding the derivative and then testing values. The derivative of this function is

\(\displaystyle f{}'(x)=3x^2+8x+5.\)

We then need to factor that and set it equal to 0, which gives us \(\displaystyle (3x+5)(x+1)=0\).

We then set each expression equal to 0 to give us our critical points.

This give us critical points at \(\displaystyle x= -\frac{5}{3}, -1\). We then set up a number line and test values on each side of the values. To the left of \(\displaystyle -\frac{5}{3}\), you can choose \(\displaystyle -2\) and plug that into the derivative. We get a positive value. In between the two critical points, you can choose \(\displaystyle -\frac{4}{3}\), which gives us a negative value. To the right of \(\displaystyle -1\), you can choose \(\displaystyle 0\), which gives a positive value. To find the minimum, you must find the point where the sign changes from negative to positive. That happens at \(\displaystyle -1\). That is the x-value of the minimum.

To find the y-value, you must plug that x-value into the original function: 

\(\displaystyle f(-1)=(-1)^3+4(-1)^2+5(-1)=-1+4-5=-2\).   

In order to solve this equation, we must first underestand that by taking the derivative of an equation of a graph and setting it equal to zero, we can find the values of \(\displaystyle x\) where there are critical points.  Critical points are either local maxima or minima, in order to figure out which you simply plug in numbers before and after that value of \(\displaystyle x\) in order to see whether or not the slope increases/decreases as is approaches or leaves that \(\displaystyle x\) value.  

In order to take the derivative of equation, the power rule must be applied, \(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\).   Taking the derivative of the graph equation, we find that it is \(\displaystyle y'=-22x^2+42\).

Setting the equation equal to zero and solving for \(\displaystyle x\), we find that the critical points for this graph are present at \(\displaystyle x=\sqrt{\frac{21}{11}},-\sqrt{\frac{21}{11}}\).  Now we must determine whether these critical points are local minima or maxima.  

In order to determine whether or not they are local minima/maxima, we must determine the slope behavior of the graph around these points.  If the slope is positive towards a critical point and negative away from that critical point, that critical point is a local maxima.  Vice versa for local minima.  \(\displaystyle x=\sqrt{\frac{21}{11}}\) is approximately 1.38.  \(\displaystyle x=-\sqrt{\frac{21}{11}}\) is approximately -1.38.  Therefore we can plug in -2,0,2 into the slope equation in order to determine the behavior of the slope around those points.

Plugging \(\displaystyle x=-2\) into the derivative of the graph equation, we find that slope is positive.

Plugging \(\displaystyle x=0\) into the derivative of the graph equation, we find that the slope is positive as well.

Plugging \(\displaystyle x=2\) into the derivative of the graph equation, we find that the slope is positive.

Because the slope is always increasing, this means that there are no local minima in this graph.