Question

attempt 1: 10 attempts remaining. given ( f(x) = e^x - 3 ), use the difference quotient in a table to estimate ( f(0.6) ). (in your calculations, do not round, but for your final answer, round to three decimal places.) ( f(0.6) = ) submit answer next item

Explanation:

Step1: Recall the difference quotient formula

The difference quotient to estimate the derivative \( f'(a) \) is given by \( f'(a) \approx \frac{f(a + h)-f(a)}{h} \) for small \( h \). We can use \( h = 0.001 \) (a small value) to get a good estimate. First, find \( f(0.6) \) and \( f(0.6 + 0.001)=f(0.601) \).

Given \( f(x)=e^{x}-3 \), so \( f(0.6)=e^{0.6}-3 \) and \( f(0.601)=e^{0.601}-3 \).

Step2: Calculate the difference quotient

The difference quotient is \( \frac{f(0.601)-f(0.6)}{0.001}=\frac{(e^{0.601}-3)-(e^{0.6}-3)}{0.001}=\frac{e^{0.601}-e^{0.6}}{0.001} \).

We know that the derivative of \( f(x)=e^{x}-3 \) is \( f'(x)=e^{x} \) (by the derivative rule of exponential function, the derivative of \( e^{x} \) is \( e^{x} \) and the derivative of a constant is 0). So we can also directly calculate \( f'(0.6)=e^{0.6} \).

Calculate \( e^{0.6} \): Using a calculator, \( e^{0.6}\approx1.82211880039 \). Rounding to three decimal places, we get \( 1.822 \).

(If we use the difference quotient method with \( h = 0.001 \)):
\( e^{0.6}\approx1.82211880039 \), \( e^{0.601}=e^{0.6 + 0.001}=e^{0.6}\times e^{0.001}\approx1.82211880039\times1.00100050017\approx1.823941417 \)
Then \( \frac{e^{0.601}-e^{0.6}}{0.001}=\frac{1.823941417 - 1.82211880039}{0.001}=\frac{1.822522617}{0.001}=1.822522617\approx1.823 \)? Wait, no, actually the exact derivative of \( e^{x} \) is \( e^{x} \), so \( f'(x)=e^{x} \), so \( f'(0.6)=e^{0.6}\approx1.822 \) (wait, let's recalculate \( e^{0.6} \) more accurately. \( 0.6=\frac{3}{5} \), \( e^{0.6}\approx e^{0.6}=1.82211880039 \), which rounds to 1.822 when rounded to three decimal places.)

Answer:

\( 1.822 \)