Question

use the formula for instantaneous rate of change, approximating the limit by using smaller and smaller values of h, to find the instantaneous rate of change for the given value. f(x)=7x^x; x = 2 the instantaneo (round to three try again. use the formula for instantaneous rate of change. find the limit by substituting progressively smaller values of h (such as 1, .01, .001) until two evaluations yield the same answer. the instantaneous rate of change is that answer. it is also possible to use the table feature on your graphing calculator to evaluate. lim h→0 (f(x + h)-f(x))/h

Explanation:

Step1: First, find \(f(x + h)\)

Given \(f(x)=7x^{x}\), then \(f(x + h)=7(x + h)^{x + h}\)

Step2: Substitute into the limit - formula

The instantaneous - rate - of - change formula is \(\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}\frac{7(x + h)^{x + h}-7x^{x}}{h}\), with \(x = 2\), we have \(\lim_{h
ightarrow0}\frac{7(2 + h)^{2 + h}-7\times2^{2}}{h}\)

Step3: Approximate the limit using small values of \(h\)

When \(h = 1\):
\[

$$\begin{align*} \frac{7(2 + 1)^{2+1}-7\times2^{2}}{1}&=\frac{7\times3^{3}-7\times4}{1}\\ &=7\times(27 - 4)\\ &=7\times23\\ &=161 \end{align*}$$

\]
When \(h=0.1\):
\[

$$\begin{align*} \frac{7(2 + 0.1)^{2 + 0.1}-7\times2^{2}}{0.1}&=\frac{7\times2.1^{2.1}-7\times4}{0.1}\\ \end{align*}$$

\]
\(2.1^{2.1}=2.1^{2}\times2.1^{0.1}\approx4.41\times1.0796=4.761036\), then \(7\times2.1^{2.1}-7\times4=7\times(4.761036 - 4)=7\times0.761036 = 5.327252\), and \(\frac{7\times2.1^{2.1}-7\times4}{0.1}=\frac{5.327252}{0.1}=53.27252\)
When \(h = 0.01\):
\[

$$\begin{align*} \frac{7(2+0.01)^{2 + 0.01}-7\times2^{2}}{0.01}&=\frac{7\times2.01^{2.01}-7\times4}{0.01} \end{align*}$$

\]
\(2.01^{2.01}=2.01^{2}\times2.01^{0.01}\approx4.0401\times1.0069=4.068977\), then \(7\times2.01^{2.01}-7\times4=7\times(4.068977 - 4)=7\times0.068977 = 0.482839\), and \(\frac{7\times2.01^{2.01}-7\times4}{0.01}=\frac{0.482839}{0.01}=48.2839\)
When \(h=0.001\):
\[

$$\begin{align*} \frac{7(2 + 0.001)^{2+0.001}-7\times2^{2}}{0.001}&=\frac{7\times2.001^{2.001}-7\times4}{0.001} \end{align*}$$

\]
\(2.001^{2.001}=2.001^{2}\times2.001^{0.001}\approx4.004001\times1.000693=4.006798\), then \(7\times2.001^{2.001}-7\times4=7\times(4.006798 - 4)=7\times0.006798=0.047586\), and \(\frac{7\times2.001^{2.001}-7\times4}{0.001}=\frac{0.047586}{0.001}=47.586\)
When \(h = 0.0001\):
\[

$$\begin{align*} \frac{7(2+0.0001)^{2 + 0.0001}-7\times2^{2}}{0.0001}&=\frac{7\times2.0001^{2.0001}-7\times4}{0.0001} \end{align*}$$

\]
\(2.0001^{2.0001}=2.0001^{2}\times2.0001^{0.0001}\approx4.00040001\times1.0000693=4.00067998\), then \(7\times2.0001^{2.0001}-7\times4=7\times(4.00067998 - 4)=7\times0.00067998 = 0.00475986\), and \(\frac{7\times2.0001^{2.0001}-7\times4}{0.0001}=\frac{0.00475986}{0.0001}=47.599\)
Rounding to three decimal places, the instantaneous rate of change is approximately \(47.600\)

Answer:

\(47.600\)