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Approximations

PROBLEM SOLVING STRATEGY: Approximations
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Problems of this type will ask you to approximate the value of a function that’s close to a value you immediately know. For instance, you might be asked to approximate $\sqrt{4.02}$, which is close to the value you know of $\sqrt{4} = 2$. Or you could be asked to approximate $\sin(0.1)$, since you know $\sin{(0)} = 0$.

Curve with red dot at (x_0, f(x_0)), and blue dot on the curve a horizontal distance delta-x away

We’ll call the point you know about $x_0$. For instance, if you want to approximate $\sqrt{4.02}$, then $x_0 = 4$. And if you want to approximate $\sin(0.1)$, then $x_0 = 0$. We represent this point that we know about, $(x_o, f(x_0))$, as the red dot in the top figure.

We’ll call the horizontal distance to the point you’re interested in $\Delta x$, so you’re looking for an approximation to $f(x_0 + \Delta x)$. We show this point as the blue dot in the lower figure. So in the examples we’re considering:
\begin{align*}
\sqrt{4.02} &= \sqrt{4 + 0.02} \text{, we have } x_0 = 4 \text{ and } \Delta x = 0.02; \\[8px] \text{and for} \\[8px] \sin(0.1) &= \sin(0 + 0.1) \text{, we have } x_0 = 0 \text{ and } \Delta x = 0.1
\end{align*}

Top figure shows the tangent line at the point (x_0, f(x_0). The lower figure shows the horizontal distance delta-x, and the associated change delta-y equals m * delta-x.

To make the approximation, we replace the function’s actual growth (or decrease) with linear growth (or decrease), which is the same thing as pretending that the function follows the tangent line as you move a small distance away from $x_0$. (See the upper figure.) Said differently, if we imagine walking from the point we know about to the point of interest in order to compute the change, instead of walking from the point along the function’s actual curve (shown in blue), we’re going to walk along the tangent line (shown in green) to get an approximation.

The method works because we can easily compute the slope $m$ of the line tangent to the curve at the point you know about, $x_0$:

$$m = \left. \frac{df}{dx} \right|_{x = x_0}$$

Then computing the approximate growth (or decrease) $\Delta y$ from the value we know about is straightforward (see the lower figure):

\begin{align*}
\Delta y &= m \, \Delta x \\
&= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x
\end{align*}

Putting all the pieces together,
\begin{align*}
f(x_0 + \Delta x) &\approx f(x_0) + \Delta y \\
&\approx f(x_0) + \left. \frac{df}{dx} \right|_{x = x_0}\Delta x
\end{align*}

This approach to estimation is known as a linear approximation since we are replacing the function around the point of interest with a line whose slope equals the tangent to the curve there.


Many students who find the discussion above rather abstract find the method straightforward after practicing the concrete problems below.

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