Question

evaluate at (h = 10^{-1}=0.1). (\frac{f(3 + 0.1)-f(3)}{0.1}=\frac{3(3.1)^{3.1}-3(3)^{3}}{0.1}) (round to three decimal places as needed.)
rewrite using the given function. simplify.
(\frac{f(3 + 1)-f(3)}{1}=\frac{3(4)^{4}-3(3)^{3}}{1}=687)
since it would be extremely difficult to simplify the limit and make it easy to evaluate, evaluate the limit by using smaller and smaller values of (h). since it is easiest to use powers of ten, begin with (h = 10^{0}=1).
(lim_{h
ightarrow0}\frac{f(3 + h)-f(3)}{h}=lim_{h
ightarrow0}\frac{3(3 + h)^{(3 + h)}-3(3)^{3}}{h})

Explanation:

Step1: Identify the function

The function seems to be \(f(x)=3x^{x}\). We first calculate \(f(3 + h)\) and \(f(3)\) for the limit \(\lim_{h
ightarrow0}\frac{f(3 + h)-f(3)}{h}\) and also for the case when \(h = 0.1\).

Step2: Calculate \(f(3 + h)\) and \(f(3)\) when \(h=0.1\)

When \(h = 0.1\), \(f(3 + 0.1)=3(3.1)^{3.1}\) and \(f(3)=3(3)^{3}\). Then \(\frac{f(3 + 0.1)-f(3)}{0.1}=\frac{3(3.1)^{3.1}-3(3)^{3}}{0.1}\).
\((3.1)^{3.1}=3.1^{\frac{31}{10}}\). Using a calculator, \(3(3.1)^{3.1}\approx3\times30.5902 = 91.7706\) and \(3(3)^{3}=3\times27 = 81\).
\(\frac{3(3.1)^{3.1}-3(3)^{3}}{0.1}=\frac{91.7706 - 81}{0.1}=\frac{10.7706}{0.1}=107.706\approx107.706\) (rounded to three - decimal places).

Step3: For the limit \(\lim_{h

ightarrow0}\frac{f(3 + h)-f(3)}{h}\)
If \(f(x)=3x^{x}\), we can use the formula \(y = x^{x}=e^{x\ln x}\), so \(f(x)=3e^{x\ln x}\).
By the definition of the derivative \(f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\). Using the chain - rule and product - rule, if \(y = 3e^{x\ln x}\), let \(u=x\ln x\), then \(\frac{du}{dx}=\ln x + 1\) and \(\frac{dy}{du}=3e^{u}\). So \(f^\prime(x)=3e^{x\ln x}(\ln x + 1)=3x^{x}(\ln x + 1)\).
Evaluating at \(x = 3\), \(f^\prime(3)=3\times3^{3}(\ln 3+ 1)=81(1.0986 + 1)=81\times2.0986=169.9866\approx169.987\)

Answer:

When \(h = 0.1\), the value is approximately \(107.706\). The value of the limit \(\lim_{h
ightarrow0}\frac{f(3 + h)-f(3)}{h}\) is approximately \(169.987\)