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 \(\displaystyle f(x)=(3x^2+x)cos(2\pi x^2)\) on the interval \(\displaystyle (0,2)\)

\(\displaystyle f(0)=(3(0)^2+0)cos(2\pi (0)^2)=0\)

\(\displaystyle f(2)=(3(2)^2+(2))cos(2\pi (2)^2)=14\)

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

\(\displaystyle \frac{14-0}{2-0}=7\)

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

Trigonometric derivative: 

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

Product rule: \(\displaystyle d[uv]=udv+vdu\)

\(\displaystyle f'(x)=(6x+1)cos(2\pi x^2)-4\pi x (3x^2+x)sin(2\pi x^2)\)

\(\displaystyle (6x+1)cos(2\pi x^2)-4\pi x (3x^2+x)sin(2\pi x^2)=7\)

Using a calculator, we find the solutions \(\displaystyle x=0.76,1,1.24,1.42,1.73\), which fit within the interval \(\displaystyle (0,2)\), satisfying the mean value theorem.

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 \(\displaystyle f(x)=(3x^2+x)cos(2\pi x^2)\) on the interval \(\displaystyle (0,2)\)

\(\displaystyle f(0)=(3(0)^2+0)cos(2\pi (0)^2)=0\)

\(\displaystyle f(2)=(3(2)^2+(2))cos(2\pi (2)^2)=14\)

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

\(\displaystyle \frac{14-0}{2-0}=7\)

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

Trigonometric derivative: 

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

Product rule: \(\displaystyle d[uv]=udv+vdu\)

\(\displaystyle f'(x)=(6x+1)cos(2\pi x^2)-4\pi x (3x^2+x)sin(2\pi x^2)\)

\(\displaystyle (6x+1)cos(2\pi x^2)-4\pi x (3x^2+x)sin(2\pi x^2)=7\)

Using a calculator, we find the solutions \(\displaystyle x=0.76,1,1.24,1.42,1.73\), which fit within the interval \(\displaystyle (0,2)\), satisfying the mean value theorem.

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 \(\displaystyle f(x)=5x^4tan(x)\) on the interval \(\displaystyle (0,0.4)\)

\(\displaystyle f(0)=5(0)^4tan(0)=0\)

\(\displaystyle f(0.4)=5(0.4)^4tan(0.4)=0.0541\)

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

\(\displaystyle \frac{0.0541-0}{0.4-0}=0.135\)

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

Trigonometric derivative: 

\(\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}\)

Product Rule: \(\displaystyle d(uv)=udv+vdu\)

\(\displaystyle f'(x)=20x^3tan(x)+\frac{5x^4}{cos^2(x)}\)

\(\displaystyle 20x^3tan(x)+\frac{5x^4}{cos^2(x)}=0.135\)

Using a calculator, we find the solution \(\displaystyle x=0.269\), which fits within the interval \(\displaystyle (0,0.4)\), satisfying the mean value theorem.

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 \(\displaystyle f(x)=tan(x)tan(2x)\) on the interval \(\displaystyle (0,0.6)\)

\(\displaystyle f(0)=tan(0)tan(2(0))=0\)

\(\displaystyle f(0.6)=tan(0.6)tan(2(0.6))=1.760\)

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

\(\displaystyle \frac{1.760-0}{0.6-0}=2.933\)

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

Trigonometric derivative: 

\(\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}\)

Product rule: \(\displaystyle d[uv]=udv+vdu\)

\(\displaystyle f'(x)=\frac{tan(2x)}{cos^2(x)}+\frac{tan(x)}{cos^2(2x)}\)

\(\displaystyle \frac{tan(2x)}{cos^2(x)}+\frac{tan(x)}{cos^2(2x)}=2.933\)

Using a calculator, we find the solutions \(\displaystyle x=0.457\), which fits within the interval \(\displaystyle (0,0.6)\), satisfying the mean value theorem.

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 \(\displaystyle f(x)=tan(x)tan(2x)\) on the interval \(\displaystyle (0,0.6)\)

\(\displaystyle f(0)=tan(0)tan(2(0))=0\)

\(\displaystyle f(0.6)=tan(0.6)tan(2(0.6))=1.760\)

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

\(\displaystyle \frac{1.760-0}{0.6-0}=2.933\)

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

Trigonometric derivative: 

\(\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}\)

Product rule: \(\displaystyle d[uv]=udv+vdu\)

\(\displaystyle f'(x)=\frac{tan(2x)}{cos^2(x)}+\frac{tan(x)}{cos^2(2x)}\)

\(\displaystyle \frac{tan(2x)}{cos^2(x)}+\frac{tan(x)}{cos^2(2x)}=2.933\)

Using a calculator, we find the solutions \(\displaystyle x=0.457\), which fits within the interval \(\displaystyle (0,0.6)\), satisfying the mean value theorem.

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 \(\displaystyle f(x)=\frac{x^4}{sin(x)}\) on the interval \(\displaystyle (0.5,0.9)\)

\(\displaystyle f(0.5)=\frac{0.5^4}{sin(0.5)}=0.1304\)

\(\displaystyle f(0.9)=\frac{0.9^4}{sin(0.9)}=0.8376\)

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

\(\displaystyle \frac{0.8376-0.1304}{0.9-0.5}=1.768\)

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

Trigonometric derivative: 

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

Quotient Rule: \(\displaystyle d[\frac{u}{v}]=\frac{vdu-udv}{v^2}\)

\(\displaystyle f'(x)=\frac{4x^3sin(x)-x^4cos(x)}{sin^2(x)}\)

\(\displaystyle \frac{4x^3sin(x)-x^4cos(x)}{sin^2(x)}=1.768\)

Using a calculator, we find the solution \(\displaystyle x=0.714\), which fits within the interval \(\displaystyle (0.5,0.9)\), satisfying the mean value theorem.

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 \(\displaystyle f(x)=ln(tan(x))\) on the interval \(\displaystyle (\frac{\pi}{4},\frac{\pi}{3})\)

\(\displaystyle f(\frac{\pi}{4})=ln(tan(\frac{\pi}{4}))=0\)

\(\displaystyle f(\frac{\pi}{3})=ln(tan(\frac{\pi}{3}))=0.549\)

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

\(\displaystyle \frac{0.549-0}{\frac{\pi}{3}-\frac{\pi}{4}}=2.097\)

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

Derivative of a natural log: 

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

Trigonometric derivative: 

\(\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}\)

\(\displaystyle f'(x)=\frac{1}{cos^2(x)tan(x)}=\frac{1}{cos(x)sin(x)}\)

\(\displaystyle \frac{1}{cos(x)sin(x)}=2.097\)

Using a calculator, we find the solution \(\displaystyle x=0.938\), which fits within the interval \(\displaystyle (\frac{\pi}{4},\frac{\pi}{3})\), satisfying the mean value theorem.

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 \(\displaystyle f(x)=ln(tan(x))\) on the interval \(\displaystyle (\frac{\pi}{4},\frac{\pi}{3})\)

\(\displaystyle f(\frac{\pi}{4})=ln(tan(\frac{\pi}{4}))=0\)

\(\displaystyle f(\frac{\pi}{3})=ln(tan(\frac{\pi}{3}))=0.549\)

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

\(\displaystyle \frac{0.549-0}{\frac{\pi}{3}-\frac{\pi}{4}}=2.097\)

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

Derivative of a natural log: 

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

Trigonometric derivative: 

\(\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}\)

\(\displaystyle f'(x)=\frac{1}{cos^2(x)tan(x)}=\frac{1}{cos(x)sin(x)}\)

\(\displaystyle \frac{1}{cos(x)sin(x)}=2.097\)

Using a calculator, we find the solution \(\displaystyle x=0.938\), which fits within the interval \(\displaystyle (\frac{\pi}{4},\frac{\pi}{3})\), satisfying the mean value theorem.

Use the power rule to find the derivative. 

\(\displaystyle \frac{d}{dx}x^3=3x^2\)

\(\displaystyle \frac{d}{dx}-3x^2=-6x\)

\(\displaystyle \frac{d}{dx}2x=2\)

Thus, the answer is \(\displaystyle 3x^2-6x+2\)

Use the power rule to find the derivative. 

\(\displaystyle \frac{d}{dx}x^3=3x^2\)

\(\displaystyle \frac{d}{dx}-3x^2=-6x\)

\(\displaystyle \frac{d}{dx}2x=2\)

Thus, the answer is \(\displaystyle 3x^2-6x+2\)