Find the derivative using the power rule. 

Remember the power rule:

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

We can now apply this to our situation.

$\displaystyle \frac{d}{dx}7x^4=28x^3$

$\displaystyle \frac{d}{dx}8x^3=24x^2$

$\displaystyle \frac{d}{dx}9x^2=18x$

$\displaystyle \frac{d}{dx}10x=10$

$\displaystyle \frac{d}{dx}1=0$

The derivative is $\displaystyle 28x^3+24x^2+18x+10$

First, find the derivative using the power rule. 

Remember the power rule:

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

We can now apply this to our situation.

$\displaystyle \frac{d}{dx}7x^2=14x$

$\displaystyle \frac{d}{dx}-4x=-4$

The derivative is $\displaystyle 14x-4$

Now, substitute $\displaystyle 6$ for $\displaystyle x$.

$\displaystyle 14(6)-4=80$

Use the product rule to find the derivative.

Remember the product rule:

$\displaystyle (f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$

We can now apply this to our situation.

$\displaystyle 5x(-\sin(x))+\cos (x)(5)$

Simplify.

$\displaystyle -5x\sin (x)+5\cos (x)$

Use the quotient rule to find the derivative. 

Remember the quotient rule:

$\displaystyle (\frac{f(x)}{g(x)})'=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}$

We can now apply this to our situation.

$\displaystyle \frac{x+2(0)-(-1)1}{(x+2)^2}$

$\displaystyle =\frac{-1}{(x+2)^2}$

Find the derivative using the quotient rule. 

Remember the quotient rule:

$\displaystyle (\frac{f(x)}{g(x)})'=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}$

We can now apply this to our situation.

$\displaystyle \frac{x^2(0)-5(2x)}{x^4}$

$\displaystyle =\frac{10x}{x^4}=\frac{10}{x^3}$

Find the derivative using the quotient rule. 

Remember the quotient rule:

$\displaystyle (\frac{f(x)}{g(x)})'=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}$

We can now apply this to our situation.

$\displaystyle \frac{d}{dx}\frac{x^2}{3}=\frac{3(2x)-x^2(0)}{9}$

$\displaystyle =\frac{6x}{9}=\frac{2x}{3}$

The mean value theorem states that for a planar arc passing through a starting and endpoint $\displaystyle (a,b);a< b$, there exists at a minimum one point, $\displaystyle c$, within the interval $\displaystyle (a,b)$ for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  $\displaystyle f(x)=x^5+7x$ on the interval $\displaystyle (0,2)$

$\displaystyle f(0)=0^5+7(0)=0$

$\displaystyle f(2)=2^5+7(2)=46$

Then take the difference of the two and divide by the interval.

$\displaystyle \frac{46-0}{2-0}=23$

Now find the derivative of the function; this will be solved for the value(s) found above.

$\displaystyle f'(x)=5x^4+7$

$\displaystyle 5x^4+7=23$

$\displaystyle x=-1.33,1.33$

The solution $\displaystyle x=1.33$ validates the mean value theorem by falling within $\displaystyle (0,2)$

The mean value theorem states that for a planar arc passing through a starting and endpoint $\displaystyle (a,b);a< b$, there exists at a minimum one point, $\displaystyle c$, within the interval $\displaystyle (a,b)$ for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of $\displaystyle f(x)=x^{10}-\frac{1}{x^{10}}$ on the interval $\displaystyle (0.9,1)$

$\displaystyle f(0.9)=0.9^{10}-\frac{1}{0.9^{10}}=-2.519$

$\displaystyle f(1)=1^{10}-\frac{1}{1^{10}}=0$ 

Then take the difference of the two and divide by the interval.

$\displaystyle \frac{0-(-2.519)}{1-0.9}=25.19$

Now find the derivative of the function; this will be solved for the value(s) found above.

$\displaystyle f'(x)=10x^9+\frac{10}{x^{11}}$

$\displaystyle 10x^9+\frac{10}{x^{11}}=25.19$

$\displaystyle x=0.94,1.09$

Of these two solutions $\displaystyle x=0.94$ validates the mean value theorem by valling within $\displaystyle (0.9,1)$

The mean value theorem states that for a planar arc passing through a starting and endpoint $\displaystyle (a,b);a< b$, there exists at a minimum one point, $\displaystyle c$, within the interval $\displaystyle (a,b)$ for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  $\displaystyle f(x)=x^3+sin(2x)+cos(3x)$ on the interval $\displaystyle (0,1)$

$\displaystyle f(0)=0^3+sin(2(0))+cos(3(0))=1$

$\displaystyle f(1)=1^3+sin(2(1))+cos(3(1))=0.919$

Then take the difference of the two and divide by the interval.

$\displaystyle \frac{0.919-1}{1-0}=-0.081$

Now find the derivative of the function; this will be solved for the value(s) found above.

$\displaystyle d(sin(u))=cos(u)du;d(cos(u))=-sin(u)du$

$\displaystyle f'(x)=3x^2+2cos(2x)-3sin(3x)$

$\displaystyle 3x^2+2cos(2x)-3sin(3x)=-0.081$

$\displaystyle x=0.247,0.811$

Both of these solutions validate the mean value theorem by falling within $\displaystyle (0,1)$.

Note that for a function that's differentiable on an interval like the one given, there will always be at least one point that satisfies the MVT.

The mean value theorem states that for a planar arc passing through a starting and endpoint $\displaystyle (a,b);a< b$, there exists at a minimum one point, $\displaystyle c$, within the interval $\displaystyle (a,b)$ for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a< c< b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  $\displaystyle f(x)=4x-sin(5x)$ on the interval $\displaystyle (0,\pi)$

$\displaystyle f(0)=4(0)-sin(5(0))=0$

$\displaystyle f(\pi)=4\pi-sin(5\pi)=4\pi$

Then take the difference of the two and divide by the interval.

$\displaystyle \frac{4\pi-0}{\pi-0}=4$

Now find the derivative of the function; this will be solved for the value(s) found above.

$\displaystyle d(sin(u))=cos(u)du$

$\displaystyle f'(x)=4-5cos(5x)$

$\displaystyle 4-5cos(5x)=4$

$\displaystyle cos(5x)=0$

$\displaystyle 5x=cos^{-1}(0)$

$\displaystyle 5x=\frac{\pi}{2}n ;n:integer$

$\displaystyle x=\frac{\pi}{10}n ;n:integer \neq 0$

Note that we do not include the start and end points as values that satisfy the mean value theorem. Therefore, there are nine values which fall within the open interval $\displaystyle (0,\pi)$ and satisfy the MVT $\displaystyle (n=[1,9])$