The equation below has a solution near x = 0. By replacing the left side of the equation by its linearization, find an approximate value for the solution. Note: Your answer should be in fraction form.

e^x+14x=7

Question

The equation below has a solution near x = 0. By replacing the left side of the equation by its linearization, find an approximate value for the solution. Note: Your answer should be in fraction form.

e^x+14x=7

anonymous · Answer

i have to say it does not have a solution that near 0 because $e^0+14\times 0=1$ not $7$

anonymous · Answer

i need the approximate value

anonymous · Answer

not the solution

anonymous · Answer

you can find the equation of the line tangent to the graph of $e^x+14x$ at $(0,1)$ but i don't see how that will give you a good approximation for the equation

anonymous · Answer

derivative of $e^x+14x-7$  is $e^x+14$
 at $0$ the slopes is $14$  and the equation of the tangent line at $(0,1)$ is $y-1=14x$

anonymous · Answer

maybe i am missing something.  are you supposed to be using newton ralphson?

anonymous · Answer

no

anonymous · Answer

i was wrong anyway, at $0$ the slope is 15

anonymous · Answer

the local linearization formula that all. well that what the chapter is about

anonymous · Answer

maybe you are supposed to be finding the zero of the line tangent to the curve of 
$$y=e^x+14x-7$$ that is easy enough, since $y'=e^x+14$ and at $0$ you get 15 for the slope. the point would be $(0,-6)$ and the equation for the tangent line would be 
$$y+6=15x$$ or 
$$y=15x-6$$
zero is therefore 
$$\frac{6}{15}$$

anonymous · Answer

a rather lousy approximation, but if that is what is needed ok.

anonymous · Answer

yea i think your right