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Snowcat [4.5K]
3 years ago
12

Use the formula for instantaneous rate of change, approximating the limit by using smaller and smaller values of hh , to find th

e instantaneous rate of change for each function at the given value.
f(x)=x^{x} at x=2
Mathematics
1 answer:
notka56 [123]3 years ago
8 0

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

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