Been trying to solve this one problem for the past 30 minutes but cannot.
The questions reads: A function is given. Determine the average rate of change of the function between the given values and the variable.
$$g(x)= \frac{1}{x}, \qquad 1 \leq x \leq a$$
My answer was $-\frac{1}{a}-1$ but that is wrong according to the book.
Please show all steps.
Best Answer
We have from the fundamental theorem of calculus that:
$$\int_{x_{1}}^{x_{2}}\frac{\mathrm{d}}{\mathrm{d}x}\left(f(x)\right)\:\mathrm{d}x=f(x_{2})-f(x_{1})$$
We also have the formula for an average of a function $f(t)$ between two values $t_{1}$ and $t_{2}$:
$$f_{\text{avg}}=\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}f(t)\:\mathrm{d}t$$
In your case, we therefore have:
$$g'_{\text{avg}}=\frac{1}{a-1}\int_{1}^{a}\frac{\mathrm{d}g(x)}{\mathrm{d}x}\:\mathrm{d}x=\frac{g(a)-g(1)}{a-1}=\frac{\frac{1}{a}-1}{a-1}=-\frac{1}{a}$$
It appears you are having trouble simplifying the fraction; note that if you multiply top and bottom by $a$ (which is valid because $\frac{a}{a}=1$) you have:
$$\frac{\frac{1}{a}-1}{a-1}=\frac{\frac{a}{a}-a}{a^{2}-a}=\frac{1-a}{a^{2}-a}=\frac{1-a}{a(a-1)}=-\frac{a-1}{a(a-1)}=-\frac{1}{a}$$