Slope and equation of a tangent at a given point
A line of slope m containing a point \left(a,b\right) has an equation of the form:
y=m\left(x-a\right)+b
The slope-intercept equation of the line with slope 2 containing the point \left(3{,}4\right) is:
y=2\left(x-3\right)+4\\y=2x-6+4\\y=2x-2
A line is uniquely determined by its slope and a point on the line.
Suppose \left(a,b\right) is a point on a line with slope m. Then:
- The graph of the linear equation f\left(x\right)=m\left(x-a\right)+b has slope m.
- f\left(a\right)=m\left(a-a\right)+b=b
Therefore, the linear equation f\left(x\right)=m\left(x-a\right)+b is the unique linear function whose graph contains the point \left(a,b\right) and has slope m.
Suppose f\left(x\right) is a function which is differentiable at x=a. Then the slope of the line tangent to the graph of f\left(x\right) at x=a is f'\left(a\right) by definition.
Consider the following function:
f\left(x\right)=x^3-4
We have:
f'\left(x\right)=3x^2
The slope of the line tangent to the graph of f\left(x\right) at x=1 is f'\left(1\right)=3.
Let f\left(x\right) be a function which is differentiable at x=a. An equation of the line tangent to the graph of f\left(x\right) at \left(a,f\left(a\right)\right) is:
y=f'\left(a\right)\left(x-a\right)+f\left(a\right)
Consider the following function f defined on \mathbb{R} :
f\left(x\right)=x^2
An equation of the line tangent to the graph of f\left(x\right) at \left(1,f\left(1\right)\right) is:
y=f'\left(1\right)\left(x-1\right)+f\left(1\right)
The derivative of f\left(x\right) is computed as follows:
f'\left(x\right)=2x
We have:
f'\left(1\right)=2\times1=2
We also have:
f\left(1\right)=1
Therefore, an equation of the line tangent to the graph of f\left(x\right)=x^2 at the point \left(1{,}1\right) is:
y=2\left(x-1\right)+1
y=2x-1
Consider the following function:
f\left(x\right)=\dfrac{1}{x-1}
An equation of the line tangent to the graph of f\left(x\right) at \left(3,f\left(3\right)\right) is:
y=f'\left(3\right)\left(x-3\right)+f\left(3\right)
The derivative of f\left(x\right) is computed as follows:
f'\left(x\right)=\left(\dfrac{1}{x-1}\right)'\\=\left(\left(x-1\right)^{-1}\right)'\\=-\left(x-1\right)^{-2}\\=\dfrac{-1}{\left(x-1\right)^2}
At x=3, the slope of the line tangent to the graph of f\left(x\right) is:
f'\left(3\right)=\dfrac{-1}{\left(3-1\right)^2}\\=\dfrac{-1}{4}
We also have:
f\left(3\right)=\dfrac{1}{3-1}=\dfrac{1}{2}
Therefore, an equation of the line tangent to the graph of f\left(x\right) at the point \left(3,\dfrac{1}{2}\right) is:
y=\dfrac{-1}{4}\left(x-3\right)+\dfrac{1}{2}
y=\dfrac{-1}{4}x+\dfrac{3}{4}+\dfrac{1}{2}
y=-\dfrac{1}{4}x+\dfrac{5}{4}
Differentiability and continuity
Let f\left(x\right) be a function which is differentiable at x=a. Then f\left(x\right) is continuous at x=a.
The trigonometric function f\left(x\right)=\sin\left(x\right) is a differentiable function with derivative f'\left(x\right)=\cos\left(x\right). Therefore, the function \sin\left(x\right) is continuous.
The converse of the above theorem is false. There are examples of continuous functions which are not differentiable.
Consider the absolute value function:
f\left(x\right)=|x|
The absolute value function is a piecewise function:
f\left(x\right)=|x|=\begin{cases} x & x\geq 0 \cr \cr -x & x\leq 0 \end{cases}
In particular:
\lim\limits_{h\to 0^+}\dfrac{f\left(0+h\right)-f\left(0\right)}{h}=\lim\limits_{h\to0^+}\dfrac{h-0}{h}=1
and:
\lim\limits_{h\to 0^-}\dfrac{f\left(0+h\right)-f\left(0\right)}{h}=\lim\limits_{h\to0^-}\dfrac{-h-0}{h}=-1
Therefore, the lefthand and righthand limits of the difference quotient of |x| do not agree when x=0 and |x| is not differentiable at x=0.
However, the absolute value function is continuous at x=0.
Mean Value Theorem
Mean Value Theorem
Let a and b be two real numbers such that a\lt b.
Suppose f\left(x\right) is a function which is continuous on a closed interval \left[a,b\right] and differentiable on the open interval \left(a,b\right). There exists some real number c such that a \lt c \lt b and:
f'\left(c\right)=\dfrac{f\left(b\right)-f\left(a\right)}{b-a}
Consider the following differentiable function:
f\left(x\right)=x^2-3x+2
The average rate of change of the function f\left(x\right) from x=2 to x=4 is:
\dfrac{f\left(4\right)-f\left(2\right)}{4-2}=\dfrac{16-12+2-\left(4-6+2\right)}{2}\\=\dfrac{10}{2}\\=5
Therefore, there exists some real number c (with 2 \lt c \lt 4 ) such that:
f'\left(c\right)=5
A car is traveling down the highway and travels a distance of 300 kilometers in two hours and thirty minutes. We can use the mean value theorem to show that at some point the car was going at least 120 kilometers per hour as follows:
Let f\left(t\right) be function that assigns the time t in hours the car has traveled to the distance covered f\left(t\right) in kilometers. Then:
- f\left(0\right)=0
- f\left(2.5\right)=300
The average rate of the function f\left(t\right) from time t=0 to time t=2.5 hours is
\dfrac{f\left(2.5\right)-f\left(0\right)}{2.5-0}=\dfrac{300}{2.5}\\=120
Therefore, the average rate of change of the car from time t=0 to time t=2.5 is 120 kilometers per hour.
By the mean value theorem, we know that there exists a real number c with 0\lt c\lt 2.5 such that:
f'\left(c\right)=120
The derivative f'\left(t\right), or the rate of change of the car traveling, is the function which measures the speed of the car with respect to time. Therefore, the car was going 120 kilometers per hour at some point.
Consider the following function:
f\left(x\right)=x^2
Observe that:
\dfrac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}=\dfrac{4-1}{3}=1
The mean value theorem guarantees the existence of some real number c such that -1\leq c\leq 2 and f'\left(c\right)=1.
This is indeed the case because f'\left(x\right)=2x is equal to 1 when x=\dfrac{1}{2} and -1\leq \dfrac{1}{2}\leq 2.
Increasing and decreasing functions
- A function is increasing on an interval if the instantaneous rate of change of the function at each point in the interval is positive.
- A function is decreasing on an interval if the instantaneous rate of change of the function at each point in the interval is negative.
Increasing and decreasing functions and their derivatives
Let f\left(x\right) be a differentiable function and a be a real number in the domain of f\left(x\right). We have the following:
- f\left(x\right) is increasing on an interval I if f'\left(a\right) \gt 0 for all a in I.
- f\left(x\right) is decreasing on an interval I if f'\left(a\right) \lt 0 for all a in I.
The following graph is the graph of f\left(x\right)=x^2. The portion of the graph in blue is where the function f\left(x\right)=x^2 has negative derivative and the portion of the graph in red is where the function f\left(x\right)=x^2 has positive derivative.
The following is the graph of a differentiable function. There are several tangent lines drawn on the graph.
The derivative of a function measures the slope of tangent lines. Therefore:
- The tangent lines with positive slope is where the function has a positive derivative.
- The tangent lines with negative slope is where the function has negative derivative.
The tangent lines which are red have positive slope and the tangent lines which are blue have negative slope.
Derivatives allow us to determine where the graph of a function is increasing and decreasing without examining the graph of the function. In order to do that, we need to follow 2 steps:
- Compute the derivative of the function
- Determine the sign of the derivative
Consider the following differentiable function:
f\left(x\right)=x^3-3x+7
The derivative of f\left(x\right) is:
3x^2-3=3\left(x^2-1\right)=3\left(x-1\right)\left(x+1\right)
The following is the sign chart of 3x^2-3 :
Therefore, the graph of f\left(x\right) has the following properties:
- f\left(x\right) is increasing on \left(-\infty, -1\right) and on \left(1,\infty\right).
- f\left(x\right) is decreasing on \left(-1{,}1\right)
The following is the graph of f\left(x\right)=x^3-3x+7.
Concavity and points of inflection
The first derivative of a function can be used to determine intervals where the graph of a function is increasing and decreasing. The second derivative of a function provides more information about the shape of a function's graph.
Concavity
- A function is concave up if the graph of the function faces upwards.
- A function is concave down if the graph of the function faces downwards.
The function f\left(x\right)=x^2 is a concave up function.
The function f\left(x\right)=-x^2 is a concave down function.
Suppose f\left(x\right) is a function twice differenciable. The graph of f\left(x\right) is said to be concave up on an interval \left(a,b\right) if and only if:
f''\left(x\right) \gt 0 for all a \lt x \lt b
Consider the following function:
f\left(x\right)=x^2
The second derivative of f\left(x\right) is:
\left(x^2\right)''=\left(2x\right)'=2
The second derivative of f\left(x\right)=x^2 is positive everywhere and so the graph of f\left(x\right) is concave up everywhere.
Suppose f\left(x\right) is a function twice differenctable. Then the graph of f\left(x\right) is said to be concave down on an interval \left(a,b\right) if:
f''\left(x\right) \lt 0 for all a \lt x \lt b
Consider the following function:
f\left(x\right)=-x^2
The second derivative of f\left(x\right) is:
\left(-x^2\right)''=\left(-2x\right)'=-2
The second derivative of f\left(x\right)=-x^2 is negative everywhere and so the graph of f\left(x\right) is concave down everywhere.
A function can have different intervals where it is concave up and concave down.
Consider the following function:
f\left(x\right)=x^4-6x^2+2x
The second derivative of f\left(x\right) is computed as follows:
f''\left(x\right)=\left(4x^3-12x+2\right)'\\=12x^2-12=12\left(x-1\right)\left(x+1\right)
The following is the sign chart of f''\left(x\right) :
Therefore, the graph of f\left(x\right) has the following properties:
- f\left(x\right) is concave up on \left(-\infty, -1\right) and on \left(1,\infty\right).
- f\left(x\right) is concave down over \left(-1{,}1\right)
The following graph is the graph of f\left(x\right)=x^4-6x^2+2x. The portion of the graph in blue is where the graph is concave up and the portion of the graph in red is where the graph is concave down.
- If a function is concave up at a point then the line tangent to the graph of the function at that point lies below the graph of the function.
- If a function is concave down at a point then the line tangent to the graph of the function at that point lies above the graph of the function.
Point of inflection
Suppose f\left(x\right) is a function. A point of inflection of f\left(x\right) is a point on the graph where the graph changes from being concave up to concave down, or vice versa.
In the graph of f\left(x\right)=x^4-6x^2+2x, the points of inflection are the points where the graph changes from blue to red and red to blue.
Suppose f\left(x\right) is a function twice differentiable. Points of inflection of f\left(x\right) are found as follows:
- Find all solutions to f''\left(x\right)=0
- Create a sign chart of f''\left(x\right)
Any point where the signs of the f''\left(x\right) change are inflection points of f\left(x\right).
Consider the following function:
f\left(x\right)=x^3
The second derivative of f\left(x\right) is:
f''\left(x\right)=\left(3x^2\right)'=6x
The solution to 6x=0 is just 0 and the following is the sign chart of f''\left(x\right) :
Therefore, the sign of f''\left(x\right) changes at x=0 and the graph of f\left(x\right)=x^3 has an inflection point at x=0.