Question

Use the formula for instantaneous rate of change, approximating the limit by using smaller and smaller values of hh , to find the instantaneous rate of change for each function at the given value.
f(x)=x^{x} at x=2

Answer 1

Answer:

$Rate = 6.7726$

Step-by-step explanation:

Given

$f(x) = x^x$ at $x =2$

Required

The instantaneous rate of change

We have:

$f(x) = x^x$

The instantaneous rate of change is:

$\lim_{h \to 0} \frac{f(a + h) -f(a)}{h}$

$x =2$ implies that: $a = 2$

So, we have:

$a = 2$      $h = 0.01$

$\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.01) -f(2)}{0.01} = \frac{f(2.01) -f(2)}{0.01} = \frac{2.01^{2.01} - 2^2}{0.01} = 6.840403$

Keep reducing h but set a constant at 2

$h = 0.001$

$\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.001) -f(2)}{0.001} = \frac{f(2.001) -f(2)}{0.001} = \frac{2.001^{2.001} - 2^2}{0.001} = 6.779327$

$h = 0.0001$

$\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.0001) -f(2)}{0.0001} = \frac{f(2.0001) -f(2)}{0.0001} = \frac{2.0001^{2.0001} - 2^2}{0.0001} = 6.773262$

$h = 0.00001$

$\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.00001) -f(2)}{0.00001} = \frac{f(2.00001) -f(2)}{0.00001} = \frac{2.00001^{2.00001} - 2^2}{0.00001} = 6.772656$

$h = 0.000001$

$\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.000001) -f(2)}{0.000001} = \frac{f(2.000001) -f(2)}{0.000001} = \frac{2.000001^{2.000001} - 2^2}{0.000001} = 6.772595$

$h = 0.0000001$

$\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.0000001) -f(2)}{0.0000001} = \frac{f(2.0000001) -f(2)}{0.0000001} = \frac{2.0000001^{2.0000001} - 2^2}{0.0000001} = 6.772589$

$h = 0.00000001$

$\frac{f(a + h) -f(a)}{h} = \frac{f(2 + 0.00000001) -f(2)}{0.00000001} = \frac{f(2.00000001) -f(2)}{0.00000001} = \frac{2.00000001^{2.00000001} - 2^2}{0.00000001} = 6.772589$

Notice that:

$\frac{f(a + h) -f(a)}{h} = 6.772589$ for $h = 0.00000001$ and $h = 0.0000001$

Hence, the instantaneous rate of change is:

$Rate = 6.772589$

$Rate = 6.7726$ ---- approximated