For the function $f(x) =xe^{x}$:

Solution (a) $dy=f^\prime (x) dx= (xe^{x}+e^{x}) dx=(x+1) e^{x}dx$.

(b) See Figure 9(a). When $x=0$ and $\Delta x=dx=0.5$, then $$\ dy=(x+1) e^{x}dx=(0+1) e^{0}(0.5) =0.5$$

Figure 9

232

The tangent line rises by $0.5$ as $x$ changes from $0$ to $0.5$. The corresponding change in the height of the graph $f$ is $$\begin{eqnarray*} \Delta y=f( x+\Delta x) -f( x) &=&f(0.5) -f( 0) =0.5e^{0.5}-0\approx 0.824\\ \vert \Delta y-dy\vert &\approx& \vert 0.824-0.5\vert =0.324 \end{eqnarray*}$$

The graph of the tangent line is approximately $0.324$ below the graph of $f$ at $x\,{=}\,0.5$.

(c) See Figure 9(b). When $x=0$ and $\Delta x=dx=0.1$, then $$dy=(0+1) e^{0}(0.1) =0.1$$

The tangent line rises by $0.1$ as $x$ changes from $0$ to $0.1.$ The corresponding change in the height of the graph $f$ is $$\begin{eqnarray*} \Delta y=f( x+\Delta x) -f( x) &=&f(0.1) -f( 0) =0.1e^{0.1}-0\approx 0.111\\ \vert \Delta y-dy\vert &\approx& \left\vert 0.111-0.1\right\vert =0.011 \end{eqnarray*}$$

The graph of the tangent line is approximately $0.011$ below the graph of $f$ at $x=0.1$.

(d) See Figure 9(c). When $x=0$ and $\Delta x=dx=0.01$, then $$dy=(0+1) e^{0}( 0.01) =0.01$$

The tangent line rises by $0.01$ as $x$ changes from $0$ to $0.01.$ The corresponding change in the height of the graph of $f$ is $$\begin{eqnarray*} \Delta y=f( x+\Delta x) -f( x) &=&f( 0.01) -f( 0) =0.01e^{0.01}-0\approx 0.0101\\ \vert \Delta y-dy\vert &\approx& \vert 0.0101-0.01\vert =0.0001 \end{eqnarray*}$$

The graph of the tangent line is approximately $0.0001$ below the graph of $f$ at $x=0.01$.

(e) The closer $\Delta x$ is to $0$, the closer $dy$ is to $\Delta y.$ So, we conclude that the closer $\Delta x$ is to $0$, the less the tangent line departs from the graph of the function. That is, we can use the tangent line to $f$ at a point $P$ as a linear approximation to $f$ near $P$.