# a) Prove, working directly from the definition, that if f(x) = 1/x, then f'(a) = -1/a^2, for a...

## Question:

a) Prove, working directly from the definition, that if {eq}f(x) = 1/x {/eq}, then {eq}f'(a) = -1/a^2 {/eq}, for {eq}a \neq 0 {/eq} .

b) Prove that the tangent line to the graph of {eq}f {/eq} at ({eq}a, 1/a {/eq}) does not intersect the graph of {eq}f {/eq}, except at ({eq}a, 1/a {/eq}) .

## Limit Definition of a Derivative

The limit definition of the derivative of a function {eq}\displaystyle f(x) {/eq} at a point {eq}\displaystyle x=a, {/eq} is

{eq}\displaystyle f'(a)= \lim_{x\to a}\frac{f(x)-f(a)}{x-a}, {/eq}

and represents the slope of the tangent line to the graph of {eq}\displaystyle f {/eq} at the point {eq}\displaystyle a. {/eq}

The tangent to the graph of {eq}\displaystyle f(x) {/eq} at the point {eq}\displaystyle a {/eq} intersect the graph at a single point of tangency, {eq}\displaystyle a, {/eq} when {eq}\displaystyle x {/eq} is very close to {eq}\displaystyle a, {/eq}

but the tangent line can intersect the graph in other points, then a.

a) To find the derivative of the function {eq}\displaystyle f(x)=\frac{1}{x} \text{ at } a, {/eq} using the limit definition as below.

{eq}\displaystyle \begin{align} f'(a)&=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\\ &=\lim_{x\to a}\frac{\dfrac{1}{x}-\dfrac{1}{a}}{x-a}\\ &=\lim_{x\to a}\frac{\dfrac{a-x}{ax}}{x-a}\\ &=\lim_{x\to a}\dfrac{-\require{\cancel}\bcancel{(x-a)}}{ax\require{\cancel}\bcancel{(x-a)}}\\ &=-\lim_{x\to a}\frac{1}{ax}=\boxed{-\frac{1}{a^2}}. \end{align} {/eq}

b. To show that the tangent line to the graph at {eq}\displaystyle f(x)=\frac{1}{x}, {/eq} at {eq}\displaystyle \left(a,\frac{1}{a}\right) {/eq}

does not intersect the graph except the point of tangency,

we will first obtain the tangent line.

The tangent line of {eq}\displaystyle y=\frac{1}{a}-\frac{1}{a^2}(x-a)\iff y=\frac{2}{a}-\frac{x}{a^2}. {/eq}

and to find the points of intersection with the graph, we will substitute y from the tangent line in the function of the graph.

{eq}\displaystyle f(x)=\frac{1}{x}\iff \frac{2}{a}-\frac{x}{a^2} =\frac{1}{x}\iff \frac{x^2-2ax+a^2}{a^2 x}=0\\ \displaystyle \iff (x-a)^2=0\iff x=a \text{ and } f(a)=\frac{1}{a} \\ \displaystyle \text{ which is the point of tangency, so this is the only point of intersection of the tangent line with the graph}. {/eq}