Question

determine the point at which the graph of the function has a horizontal tangent line.
f(x)=\frac{2x - 3}{x^{2}}
((x,y)=(quad))

Explanation:

Step1: Use quotient - rule for differentiation

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 2x - 3$, $u^\prime=2$, $v = x^{2}$, and $v^\prime = 2x$. So, $f^\prime(x)=\frac{2\cdot x^{2}-(2x - 3)\cdot2x}{(x^{2})^{2}}$.

Step2: Simplify the derivative

\[

$$\begin{align*} f^\prime(x)&=\frac{2x^{2}-4x^{2}+6x}{x^{4}}\\ &=\frac{- 2x^{2}+6x}{x^{4}}\\ &=\frac{-2x(x - 3)}{x^{4}}\\ &=\frac{-2(x - 3)}{x^{3}},x eq0 \end{align*}$$

\]

Step3: Set the derivative equal to zero

A horizontal tangent line occurs when $f^\prime(x)=0$. So, $\frac{-2(x - 3)}{x^{3}}=0$. For a fraction to be zero, the numerator must be zero while the denominator is non - zero. Set $-2(x - 3)=0$. Solving $x - 3=0$ gives $x = 3$.

Step4: Find the y - coordinate

Substitute $x = 3$ into the original function $f(x)=\frac{2x-3}{x^{2}}$. Then $f(3)=\frac{2\times3 - 3}{3^{2}}=\frac{6 - 3}{9}=\frac{3}{9}=\frac{1}{3}$.

Answer:

$(3,\frac{1}{3})$