Question

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estimate the rate of change of the function at the given point. 6. ( f(x) = sqrt{2x} ) at ( x = 3 ) hwi.2a # 6,7,8,9

Explanation:

Step1: Recall the rate of change formula

The average rate of change of a function \( f(x) \) near \( x = a \) can be approximated by the difference quotient \(\frac{f(a + h)-f(a)}{h}\) for a small \( h \). Let's take \( h = 0.001 \) (a small value) to approximate the instantaneous rate of change (which is the derivative at \( x = 3 \)). First, find \( f(3) \) and \( f(3 + 0.001)=f(3.001) \).

The function is \( f(x)=\sqrt{2x} \), so \( f(3)=\sqrt{2\times3}=\sqrt{6}\approx 2.4495 \)

\( f(3.001)=\sqrt{2\times3.001}=\sqrt{6.002}\approx 2.450 \)

Step2: Calculate the difference quotient

Using the difference quotient \(\frac{f(3.001)-f(3)}{0.001}\), substitute the values:

\(\frac{2.450 - 2.4495}{0.001}=\frac{0.0005}{0.001} = 0.5\)

Alternatively, we can find the derivative of \( f(x) \). The function \( f(x)=\sqrt{2x}=(2x)^{\frac{1}{2}} \). Using the power rule and chain rule, \( f^\prime(x)=\frac{1}{2}(2x)^{-\frac{1}{2}}\times2=\frac{1}{\sqrt{2x}} \)

At \( x = 3 \), \( f^\prime(3)=\frac{1}{\sqrt{2\times3}}=\frac{1}{\sqrt{6}}\approx\frac{1}{2.4495}\approx 0.4082 \) (Wait, there was a mistake in the first approximation with \( h = 0.001 \), let's recalculate \( f(3.001) \) more accurately. \( 2\times3.001 = 6.002 \), \(\sqrt{6.002}\approx\sqrt{6 + 0.002}\). Using linear approximation: Let \( y=\sqrt{u} \), \( u = 6 \), \( \Delta u=0.002 \), \( y^\prime=\frac{1}{2\sqrt{u}} \), so \( \Delta y\approx\frac{1}{2\sqrt{6}}\times0.002 \), \( y+\Delta y=\sqrt{6}+\frac{0.002}{2\sqrt{6}}\approx 2.4495+\frac{0.001}{2.4495}\approx 2.4495 + 0.000408\approx 2.449908 \)

Then \(\frac{f(3.001)-f(3)}{0.001}=\frac{2.449908 - 2.4495}{0.001}=\frac{0.000408}{0.001}=0.408\), which matches the derivative result.

The derivative method is exact for the instantaneous rate of change. So the rate of change of \( f(x)=\sqrt{2x} \) at \( x = 3 \) is \( f^\prime(3)=\frac{1}{\sqrt{6}}=\frac{\sqrt{6}}{6}\approx 0.408 \)

Answer:

The rate of change of \( f(x)=\sqrt{2x} \) at \( x = 3 \) is \(\frac{\sqrt{6}}{6}\) (or approximately \( 0.408 \))