Use a graphing calculator or computer to verify the given linear approximation at a = 0? Then determine the values of x for which the linear approximation is accurate to within 0.1. (Round the answers to three decimal places.) e^x ~= 1 + x

See the explanation section.

Here are ##y=e^x## and ##y=1+x## on the same coordinate system near ##0##. (You can scroll in or out and drag the graph around with a mouse.)

graph{(y-e^x)(y-(1+x))=0 [-5.723, 6.763, -2.37, 3.873]}

We can see that the curves are close to each other.

To answer the second question we can graph the difference,

##y=e^x-(1+x)## and try to determine when ##y## is between ##-0.1## and ##0.1##

graph{y=e^x-(1+x) [-1.085, 1.049, -0.406, 0.66]}

I prefer to graph ##y = [e^x-(1+x)]-0.1## . The difference between ##e^x## and ##1+x## is less than ##0.1## when this ##y## is negative. So we just find the ##x## intercepts: approximately ##-0.483## and ##0.416##

graph{y = e^x-(1+x)-0.1 [-1.923, 1.922, -0.959, 0.963]}