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 $\displaystyle (n=[1,9])$

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 $\displaystyle f(x)=e^x+sin(3x)$ on the interval $\displaystyle (0,\pi)$

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

$\displaystyle f(\pi)=e^\pi+sin(3\pi)=e^\pi$

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

$\displaystyle \frac{e^{\pi}-1}{\pi-0}=7.048$

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

$\displaystyle d(e^u)=e^udu$

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

$\displaystyle f'(x)=e^x+3cos(3x)$

$\displaystyle e^x+3cos(3x)=7.048$

$\displaystyle x=1.732$

There is only one value within the open interval $\displaystyle (0,\pi)$ that satisfies the mean value theorem, but there is still a value regardless.

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function$\displaystyle f(x)=\frac{x^5}{5}+\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^2}{2}+x+1$ at $\displaystyle x=1.2$

The slope of the tangent is

$\displaystyle f'(x)=x^4+x^3+x^2+x+1$

$\displaystyle f'(1.2)=1.2^4+1.2^3+1.2^2+1.2+1=7.44$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$\displaystyle -0.13$

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function $\displaystyle f(x)=x^3-2x^2+x+4$ at $\displaystyle x=2$

The slope of the tangent is

$\displaystyle f'(x)=3x^2-4x+1$

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

A line can be written as a function of the form $\displaystyle y=mx+b;m:slope,b:y-intercept$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$\displaystyle y=-\frac{1}{5}x+b$

Now to find this next constant, the normal line and the orignal function should intercept at $\displaystyle x=2$

$\displaystyle f(2)=2^3-2(2)^2+2+4=6$

$\displaystyle -\frac{2}{5}+b=6$

$\displaystyle b=\frac{32}{5}$

$\displaystyle y=-\frac{1}{5}x+\frac{32}{5}$

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 $\displaystyle f(x)=x^9+\frac{1}{x^9}$ on the interval $\displaystyle (1,2)$

$\displaystyle f(1)=1^9+\frac{1}{1^9}=2$

$\displaystyle f(2)=2^9+\frac{1}{2^9}=512.002$

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

$\displaystyle \frac{512.002-2}{2-1}=510.002$

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

$\displaystyle f'(x)=9x^8-\frac{9}{x^{10}}$

$\displaystyle 9x^8-\frac{9}{x^{10}}=510.002$

$\displaystyle x=-1.6565,1.6565$

Of these solutions, $\displaystyle x=1.6565$ satisfies the mean value theorem by falling within $\displaystyle (1,2)$

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

$\displaystyle d[a^u]=a^uduln(a)$

Trigonometric derivative: 

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

Taking the derivative of the function $\displaystyle f(x)=cos(1.2^{x^{1.3}})$ at $\displaystyle x=1.4$

The slope of the tangent is

$\displaystyle f'(x)=-1.3x^{0.3}(1.2^{x^{1.3}}ln(1.2))sin(1.2^{x^{1.3}})$

$\displaystyle f'(1.4)=-1.3(1.4)^{0.3}(1.2^{1.4^{1.3}}ln(1.2))sin(1.2^{1.4^{1.3}})=-0.34$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$\displaystyle 2.96$

To compute the derivative, the following rules will need to be applied:

 $\displaystyle \frac{d}{dx}[e^u]=e^u\frac{du}{dx}$

 AND

Product Rule:

$\displaystyle \frac{d}{dx}uv]=uv'+vu'$

Where u & v are differentiable functions.

Applying these rules:

$\displaystyle f'(x)=3e^{2x^2\sin(2x)}\frac{d}{dx}[2x^2\sin(2x)]$

$\displaystyle f'(x)=3e^{2x^2\sin(2x)}[(2x^2)(\cos(2x))(2)+(4x)(\sin(2x))]$

$\displaystyle f'(x)=3e^{2x^2\sin(2x)}[4x^2\cos(2x)+4x\sin(2x)]$

We can factor out the common 4x and result in the following:

$\displaystyle 12xe^{2x^2\sin(2x)}[x\cos(2x)+\sin(2x)]$

This is one of the answer choices.

To compute the derivative, understand that the derivative of the natural logarithm is found by:

$\displaystyle \frac{d}{dx}[\ln(u)]=\frac{1}{u}\frac{du}{dx}$

Applying this results in:

$\displaystyle f'(x)=\frac{1}{\sin(3x+8e^{2x})}\frac{d}{dx}[\sin(3x+8e^{2x})]$

$\displaystyle f'(x)=\frac{1}{\sin(3x+8e^{2x})}[\cos(3x+8e^{2x})]\frac{d}{dx}[3x+8e^{2x}]$

$\displaystyle f'(x)=\frac{1}{\sin(3x+8e^{2x})}[\cos(3x+8e^{2x})][3+8e^{2x}\frac{d}{dx}[2x]]$

$\displaystyle f'(x)=\frac{1}{\sin(3x+8e^{2x})}(\cos(3x+8e^{2x}))(3+16e^{2x})$

After some rearranging:

$\displaystyle f'(x)=\frac{(3+16e^{2x})\cos(3x+8e^{2x})}{\sin(3x+8e^{2x})}$

Notice that the cosine & sine function all the same elements within the parenthesis. We can use the following identity to rewrite this function:

$\displaystyle \frac{\cos(u)}{\sin(u)}=\cot(u)$

The function can now be rewritten as:

$\displaystyle f'(x)=(3+16e^{2x})\cot(3x+8e^{2x})$

This is one of the answer choices.