Question

a function (f) and value (a) are given below.(f(x)=\frac{2}{x + 9}), (a = 5)approximate the limit of the difference quotient,(lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}),using (h=pm0.1,pm0.01). (round your answer to within three decimal places if necessary, but do not round until your final computation.)a. when (h = 0.1), (\frac{f(a + h)-f(a)}{h}=)b. when (h=-0.1), (\frac{f(a + h)-f(a)}{h}=)c. when (h = 0.01), (\frac{f(a + h)-f(a)}{h}=)d. when (h=-0.01), (\frac{f(a + h)-f(a)}{h}=)

Explanation:

Step1: Find \(f(a)\)

Given \(f(x)=\frac{2}{x + 9}\) and \(a = 5\), then \(f(a)=f(5)=\frac{2}{5+9}=\frac{2}{14}=\frac{1}{7}\)

Step2: General formula for \(f(a + h)\)

\(f(a + h)=\frac{2}{(a + h)+9}=\frac{2}{(5 + h)+9}=\frac{2}{h + 14}\)

Step3: Calculate \(\frac{f(a + h)-f(a)}{h}\)

\(\frac{f(a + h)-f(a)}{h}=\frac{\frac{2}{h + 14}-\frac{1}{7}}{h}=\frac{\frac{14-(h + 14)}{7(h + 14)}}{h}=\frac{\frac{-h}{7(h + 14)}}{h}=-\frac{1}{7(h + 14)}\)

Step4a: When \(h = 0.1\)

Substitute \(h = 0.1\) into \(-\frac{1}{7(h + 14)}\), we get \(-\frac{1}{7(0.1+14)}=-\frac{1}{7\times14.1}=-\frac{1}{98.7}\approx - 0.010\)

Step4b: When \(h=-0.1\)

Substitute \(h=-0.1\) into \(-\frac{1}{7(h + 14)}\), we get \(-\frac{1}{7(-0.1 + 14)}=-\frac{1}{7\times13.9}=-\frac{1}{97.3}\approx - 0.010\)

Step4c: When \(h = 0.01\)

Substitute \(h = 0.01\) into \(-\frac{1}{7(h + 14)}\), we get \(-\frac{1}{7(0.01+14)}=-\frac{1}{7\times14.01}=-\frac{1}{98.07}\approx - 0.010\)

Step4d: When \(h=-0.01\)

Substitute \(h=-0.01\) into \(-\frac{1}{7(h + 14)}\), we get \(-\frac{1}{7(-0.01+14)}=-\frac{1}{7\times13.99}=-\frac{1}{97.93}\approx - 0.010\)

Answer:

a. \(-0.010\)
b. \(-0.010\)
c. \(-0.010\)
d. \(-0.010\)