Chapter 12.3, Problem 59E

To determine

ToEvaluate:Thederivative of $f\left(x\right)=x^2-4x+3$ . Use the derivative to determine any points on the graph of $f$ at which the tangent line is horizontal and use a graphing utility to verify the result.

Answer to Problem 59E

The derivative of $f\left(x\right)=x^2-4x+3$ is $f\text{'}\left(x\right)=2x-4$

The point on the graph of $f$ at which the tangent line is horizontal is $\left(x,y\right)=\left(2,-1\right)$

From the graph of $f$ and the tangent line, it is verified that the point on the graph of $f$ at which the tangent line is horizontal is $\left(x,y\right)=\left(2,-1\right)$

Explanation of Solution

Given:

The function $f\left(x\right)=x^2-4x+3$

Concept Used:

The derivative of $f$ at $x$ is given by

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ provided the limit exists.

The tangent line is horizontal when the slope of tangent line is zero.

The equation of the tangent line to the graph of a function $f$ at the point $\left(a,b\right)$ with slope $m$ is given by

  $y-b=m\left(x-a\right)$

  ${\left(a+b\right)}^2=a^2+b^2+2ab$

Calculation:

Forthe function $f\left(x\right)=x^2-4x+3$

The derivative of $f$ at $x$ is given by

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ provided the limit exists.

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\left({\left(x+h\right)}^2-4\left(x+h\right)+3\right)-\left(x^2-4x+3\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\left(x^2+h^2+2xh-4x-4h+3\right)-\left(x^2-4x+3\right)}{h}\right){\left(a+b\right)}^2=a^2+b^2+2ab\end{array}$

  $\begin{array}{c}=\underset{h\to 0}{\lim }\left(\frac{x^2+h^2+2xh-4x-4h+3-x^2+4x-3}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h^2+2xh-4h}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h\left(h+2x-4\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(h+2x-4\right)\end{array}$

  $\begin{array}{c}=0+2x-4\text{ Substituting limit}\\ =2x-4\end{array}$

Therefore, the derivative of $f\left(x\right)=x^2-4x+3$ is $f\text{'}\left(x\right)=2x-4$

The tangent line is horizontal when the slope of tangent line is zero.

Since the derivative of a function is the slope of its tangent line.

Substituting the derivative of $f$ or the slope of tangent equal to zero, we have

  $f\text{'}\left(x\right)=2x-4=0$

  $\begin{array}{c}2x-4=0\\ 2x=4\\ x=\frac{4}{2}\\ x=2\end{array}$

Substituting in the function $y=f\left(x\right)=x^2-4x+3$

  $\begin{array}{c}y={\left(2\right)}^2-4\left(2\right)+3\\ =4-8+3\\ =-4+3\\ =-1\end{array}$

Therefore, the point on the graph of $f$ at which the tangent line is horizontal is $\left(x,y\right)=\left(2,-1\right)$

For the equation of tangent line

Since, the tangent line is horizontal, so the slope of tangent line is $m=0$

The point on the tangent line is $\left(x,y\right)=\left(2,-1\right)$

The equation of the tangent line to the graph of a function $f$ at the point $\left(a,b\right)$ with slope $m$ is given by

  $y-b=m\left(x-a\right)$

  $\begin{array}{c}\left(y-\left(-1\right)\right)=0\left(x-2\right)\\ y+1=0\\ y=-1\end{array}$

Plotting tangent line and the graph of $f$ , on the same graph using a graphing utility, we have

  

Therefore, from the above graph, it is verified that the point on the graph of $f$ at which the tangent line is horizontal is $\left(x,y\right)=\left(2,-1\right)$

Conclusion:

Thederivative of $f\left(x\right)=x^2-4x+3$ is $f\text{'}\left(x\right)=2x-4$

The point on the graph of $f$ at which the tangent line is horizontal is $\left(x,y\right)=\left(2,-1\right)$

From the graph of $f$ and the tangent line, it is verified that the point on the graph of $f$ at which the tangent line is horizontal is $\left(x,y\right)=\left(2,-1\right)$