Question #6f5b6

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Question #6f5b6

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Answer 1 · 2015-09-14T14:32:38

See the explanation section.

Explanation:

Here are $y = e^{x}$ $y=e^x$ and $y = 1 + x$ $y=1+x$ on the same coordinate system near $0$ $0$ . (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,

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

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

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

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

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Question #6f5b6

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