12.4 Derivatives  (Page 9/18)

How is the slope of a linear function similar to the derivative?

What is the difference between the average rate of change of a function on the interval $\left[x,x+h\right]$ and the derivative of the function at $x?$

A car traveled 110 miles during the time period from 2:00 P.M. to 4:00 P.M. What was the car's average velocity? At exactly 2:30 P.M. , the speed of the car registered exactly 62 miles per hour. What is another name for the speed of the car at 2:30 P.M. ? Why does this speed differ from the average velocity?

Explain the concept of the slope of a curve at point $x.$

Suppose water is flowing into a tank at an average rate of 45 gallons per minute. Translate this statement into the language of mathematics.

$f\left(x\right)=3x-4$

$f\left(x\right)=-2x+1$

$f\left(x\right)=x^2-2x+1$

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

$f\left(x\right)=2x^2+5$

$f\left(x\right)=\frac{-1}{x-2}$

$f\left(x\right)=\frac{2+x}{1-x}$

$f\left(x\right)=\frac{5-2x}{3+2x}$

$f\left(x\right)=\sqrt{1+3x}$

$f(x)=3x^3-x^2+2x+5$

$f(x)=5$

$f(x)=5\pi$

$\left(\mathrm{-2},0\right)$ and $\left(\mathrm{-4},5\right)$

$\left(4,\mathrm{-3}\right)$ and $\left(\mathrm{-2},\mathrm{-1}\right)$

$\left(0,5\right)$ and $\left(6,5\right)$

$\left(7,\mathrm{-2}\right)$ and $\left(7,10\right)$

$f(x)=x^3+1$

$f(x)=-3x^2-7x=6$

$f(x)=7x^2$

$f(x)=3x^3+2x^2+x-26$

$\begin{array}{ll}f(x)=2x^2-3x\hfill & x=3\hfill \end{array}$

$\begin{array}{ll}f(x)=x^3+1\hfill & x=2\hfill \end{array}$

$\begin{array}{ll}f(x)=\sqrt{x}\hfill & x=9\hfill \end{array}$

$\begin{array}{ll}f(x)=x^2-kx,\hfill & y=4x-9\hfill \end{array}$

$f\left(-1\right)$

$f\left(0\right)$

$f\left(1\right)$

$f\left(2\right)$

$f(3)$

$f^{\prime }\left(-1\right)$

$f^{\prime }\left(0\right)$

$f^{\prime }(1)$

$f^{\prime }\left(2\right)$

$f^{\prime }\left(3\right)$

Sketch the function based on the information below:

$f^{\prime }\left(x\right)=2x$ , $f\left(2\right)=4$

Numerically evaluate the derivative. Explore the behavior of the graph of $f(x)=x^2$ around $x=1$ by graphing the function on the following domains: $\left[0.9,1.1\right]$ , $\left[0.99,1.01\right]$ , $\left[0.999,1.001\right],$ and $[0.9999,1.0001]$ . We can use the feature on our calculator that automatically sets Ymin and Ymax to the Xmin and Xmax values we preset. (On some of the commonly used graphing calculators, this feature may be called ZOOM FIT or ZOOM AUTO). By examining the corresponding range values for this viewing window, approximate how the curve changes at $x=1,$ that is, approximate the derivative at $x=1.$