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 function at the given value.
f(x)=5x^x; x = 3

the instantaneous rate of change for the function at x = 3 is .
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for instantaneous rate of change

The instantaneous rate of change of a function $y = f(x)$ at $x=a$ is given by $\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $f(x)=5x^{x}$, $a = 3$. So we need to approximate $\lim_{h
ightarrow0}\frac{5(3 + h)^{3 + h}-5\times3^{3}}{h}$.

Step2: First, find $f(3)$

$f(3)=5\times3^{3}=5\times27 = 135$.

Step3: Approximate the limit using small - values of $h$

Let's start with $h = 0.001$.
$f(3 + 0.001)=5\times(3+0.001)^{3 + 0.001}=5\times3.001^{3.001}$.
Using a calculator, $3.001^{3.001}=3^{3}\times3.001^{0.001}\approx27\times(1 + 0.001\ln(3.001))\approx27\times(1+0.001\times1.099)=27\times1.001099 = 27.029673$.
So $f(3 + 0.001)=5\times27.029673=135.148365$.
Then $\frac{f(3 + 0.001)-f(3)}{0.001}=\frac{135.148365 - 135}{0.001}=\frac{0.148365}{0.001}=148.365$.
Let's take $h=- 0.001$.
$f(3-0.001)=5\times(3 - 0.001)^{3-0.001}=5\times2.999^{2.999}$.
$2.999^{2.999}=2.999^{3}\times2.999^{-0.001}\approx27\times(1-0.001\ln(2.999))\approx27\times(1 - 0.001\times1.0986)=27\times0.9989014 = 26.9703378$.
So $f(3 - 0.001)=5\times26.9703378 = 134.851689$.
Then $\frac{f(3-0.001)-f(3)}{-0.001}=\frac{134.851689 - 135}{-0.001}=\frac{-0.148311}{-0.001}=148.311$.
Averaging these two approximations: $\frac{148.365+148.311}{2}=148.338$.

Answer:

$148.338$