Answer:
[tex]Rate = 6.7726[/tex]
Step-by-step explanation:
Given
[tex]f(x) = x^x[/tex] at [tex]x =2[/tex]
Required
The instantaneous rate of change
We have:
[tex]f(x) = x^x[/tex]
The instantaneous rate of change is:
[tex]\lim_{h \to 0} \frac{f(a + h) -f(a)}{h}[/tex]
[tex]x =2[/tex] implies that: [tex]a = 2[/tex]
So, we have:
[tex]a = 2[/tex] [tex]h = 0.01[/tex]
[tex]\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.01) -f(2)}{0.01} = \frac{f(2.01) -f(2)}{0.01} = \frac{2.01^{2.01} - 2^2}{0.01} = 6.840403[/tex]
Keep reducing h but set a constant at 2
[tex]h = 0.001[/tex]
[tex]\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.001) -f(2)}{0.001} = \frac{f(2.001) -f(2)}{0.001} = \frac{2.001^{2.001} - 2^2}{0.001} = 6.779327[/tex]
[tex]h = 0.0001[/tex]
[tex]\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.0001) -f(2)}{0.0001} = \frac{f(2.0001) -f(2)}{0.0001} = \frac{2.0001^{2.0001} - 2^2}{0.0001} = 6.773262[/tex]
[tex]h = 0.00001[/tex]
[tex]\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.00001) -f(2)}{0.00001} = \frac{f(2.00001) -f(2)}{0.00001} = \frac{2.00001^{2.00001} - 2^2}{0.00001} = 6.772656[/tex]
[tex]h = 0.000001[/tex]
[tex]\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.000001) -f(2)}{0.000001} = \frac{f(2.000001) -f(2)}{0.000001} = \frac{2.000001^{2.000001} - 2^2}{0.000001} = 6.772595[/tex]
[tex]h = 0.0000001[/tex]
[tex]\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.0000001) -f(2)}{0.0000001} = \frac{f(2.0000001) -f(2)}{0.0000001} = \frac{2.0000001^{2.0000001} - 2^2}{0.0000001} = 6.772589[/tex]
[tex]h = 0.00000001[/tex]
[tex]\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.00000001) -f(2)}{0.00000001} = \frac{f(2.00000001) -f(2)}{0.00000001} = \frac{2.00000001^{2.00000001} - 2^2}{0.00000001} = 6.772589[/tex]
Notice that:
[tex]\frac{f(a + h) -f(a)}{h} = 6.772589[/tex] for [tex]h = 0.00000001[/tex] and [tex]h = 0.0000001[/tex]
Hence, the instantaneous rate of change is:
[tex]Rate = 6.772589[/tex]
[tex]Rate = 6.7726[/tex] ---- approximated
