Rolle's Theorem states that if a real-valued function, \(\displaystyle f(x)\), is continuous and differentiable on a closed interval \(\displaystyle [a,b]\), and if\(\displaystyle f(a)=f(b)\), then there must be some point, \(\displaystyle c\), in the open interval \(\displaystyle (a,b)\) that satisfies the equation \(\displaystyle f'(c)=0\).

Essentially, if a function has a start and end point value which are equivalent, then assuming that the function has no breaks or abrupt and unsmooth changes in slope, there must be points where the derivative is zero, the slope is flat, and there is no change in the function value.

Visually \(\displaystyle f(x) =x^2-4x+2\) is continous over \(\displaystyle [1,3]\)

Evaluate the function \(\displaystyle f(x) =x^2-4x+2\)

at its specified start and end points \(\displaystyle [1,3]\)

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

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

This satisfies \(\displaystyle f(a)=f(b)\) for Rolle's Theorem. Now differentiate:

\(\displaystyle f'(x)=2x-4\)

The function is differentiable over the interval because its derivative is continuous; Rolle's Theorem is fully satisfied. 

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

\(\displaystyle c=2\)

Rolle's Theorem states that if a real-valued function, \(\displaystyle f(x)\), is continuous and differentiable on a closed interval \(\displaystyle [a,b]\), and if\(\displaystyle f(a)=f(b)\), then there must be some point, \(\displaystyle c\), in the open interval \(\displaystyle (a,b)\) that satisfies the equation \(\displaystyle f'(c)=0\).

Essentially, if a function has a start and end point value which are equivalent, then assuming that the function has no breaks or abrupt and unsmooth changes in slope, there must be points where the derivative is zero, the slope is flat, and there is no change in the function value.

\(\displaystyle f(x)=x^2-5x+8\) is continuous from visual inspection. Evaluating the start and end points specified by \(\displaystyle [2,4]\)

\(\displaystyle f(2)=2^2-5(2)+8=2\)

\(\displaystyle f(4)=4^2-5(4)+8=4\)

\(\displaystyle f(2)\neq f(4)\)

Rolle's Theorem is not satisfied.

Rolle's Theorem states that if a real-valued function, \(\displaystyle f(x)\), is continuous and differentiable on a closed interval \(\displaystyle [a,b]\), and if\(\displaystyle f(a)=f(b)\), then there must be some point, \(\displaystyle c\), in the open interval \(\displaystyle (a,b)\) that satisfies the equation \(\displaystyle f'(c)=0\).

Essentially, if a function has a start and end point value which are equivalent, then assuming that the function has no breaks or abrupt and unsmooth changes in slope, there must be points where the derivative is zero, the slope is flat, and there is no change in the function value.

The function \(\displaystyle f(x)=|2x|\) has no discontinuities, and looking at start/end points for the interval \(\displaystyle [-1,1]\):

\(\displaystyle f(-1)=|2(-1)|=2\)

\(\displaystyle f(1)=|2(1)|=2\)

we see that \(\displaystyle f(-1)=f(1)\).

However, the question becomes whether or not the function is differentiable over this interval.

\(\displaystyle f'(x)=-2;x< 0\)

\(\displaystyle f'(x)=2;x>0\)

The limit of \(\displaystyle f'(x)\) as \(\displaystyle x\rightarrow0\) is different depending on if we approach zero from the left or the right. The derivative is not continuous, and the function is not differentiable over \(\displaystyle [-1,1]\).

Rolle's Theorem is not satisfied.

Rolle's Theorem states that if a real-valued function, \(\displaystyle f(x)\), is continuous and differentiable on a closed interval \(\displaystyle [a,b]\), and if\(\displaystyle f(a)=f(b)\), then there must be some point, \(\displaystyle c\), in the open interval \(\displaystyle (a,b)\) that satisfies the equation \(\displaystyle f'(c)=0\).

Essentially, if a function has a start and end point value which are equivalent, then assuming that the function has no breaks or abrupt and unsmooth changes in slope, there must be points where the derivative is zero, the slope is flat, and there is no change in the function value.

Looking at the function \(\displaystyle f(x)=|x^3|\), there are no obvious discontinuities. Now consider its start and end points \(\displaystyle [-2,2]\)

\(\displaystyle f(-2)=|(-2)^3|=8\)

\(\displaystyle f(2)=|2^3|=8\)

\(\displaystyle f(-2)=f(2)\)

So far so good.

Now, the question is whether or not the function is differentiable.

\(\displaystyle f(x)=-x^3;x< 0\)

\(\displaystyle f(x)=0;x=0\)

\(\displaystyle f(x)=x^3;x>0\)

Compare the derivative of these three functions at \(\displaystyle x=0\)

\(\displaystyle -3x^2;-3(0^2)=0\)

\(\displaystyle 0\)

\(\displaystyle 3x^2;3(0^2)=0\)

The function is differentiable as well. All of the criteria for Rolle's Theorem are satisfied. And, as it stands, there is just one point that satisfies \(\displaystyle f'(c)=0\), and we just evaluated it.

\(\displaystyle c=0\)

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 \(\displaystyle f(x)=x^3-5x+7\) at the point \(\displaystyle x=2\)

The slope of the tangent is

 \(\displaystyle f'(x)=3x^2-5\)

\(\displaystyle f'(2)=3(2)^2-5=7\)

The equation of a tangent line follows the form

\(\displaystyle y=mx+b\), where m is the slope (just found), and b is a constant to ensure the line intercepts the original function.

So we currently have

\(\displaystyle y=7x+b\)

\(\displaystyle f(2)=(2)^3-5(2)+7=5\)

Since this is where the tangent must intercept the function at \(\displaystyle x=2\):

\(\displaystyle 7(2)+b=5\)

\(\displaystyle b=-9\)

\(\displaystyle y=7x-9\)

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: 

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

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function \(\displaystyle f(x)=x^3+2^{3x}\) at \(\displaystyle x=2\).

The slope of the tangent is

 \(\displaystyle f'(x)=3x^2+3(2^{3x})ln(2)\)

\(\displaystyle f'(2)=3(2)^2+3(2^{3(2)})ln(2)=145.1\)

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: 

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

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

Note that u and v may represent large functions, and not just individual variables!

Taking the derivative of the function \(\displaystyle f(x)=x^22^x\) at \(\displaystyle x=3\).

The slope of the tangent is

 \(\displaystyle f'(x)=2x2^{x}+x^22^{x}ln(2)\)

\(\displaystyle f'(3)=2(3)2^{3}+3^{2}2^{3}ln(2)=97.9\)

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 \(\displaystyle f(x)=x^6+x^5+x^4+x^3+x^2+x+1\) at \(\displaystyle x=1\).

The slope of the tangent is

\(\displaystyle f'(x)=6x^5+5x^4+4x^3+3x^2+2x+1\)

\(\displaystyle f'(1)=6(1)^5+5(1)^4+4(1)^3+3(1)^2+2(1)+1=21\)

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: 

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

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

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function \(\displaystyle f(x)=cos(3^{x})\) at the point \(\displaystyle x=\pi\).

The slope of the tangent is

\(\displaystyle f'(x)=-3^{x}sin(3^{x})ln(3)\)

\(\displaystyle f'(\pi)=-3^{\pi}sin(3^{\pi})ln(3)=-4.44\)

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: 

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

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

Note that u and v may represent large functions, and not just individual variables!

Taking the derivative of the function\(\displaystyle f(x)=\frac{4^{x}}{cos(x)}\) at \(\displaystyle x=\pi\).

The slope of the tangent is

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

\(\displaystyle f'(\pi)=\frac{4^{\pi}cos(\pi)ln(4)+4^{\pi}sin(\pi)}{cos^2(\pi)}=-107.96\)