PLEASE HELP the approximate value of y=sqrt(3+e^x) at x=0.08, obtained from the tangent to the graph at x=0, is 1) 2.01 2) 2.02 3) 2.03 4) 2.04 5) 2.05 explanation would be greatly appreciated
this is "linear approximation." Have you used this method before? You are asked to approximate the value of y=sqrt(3+e^x) at x=0.08. Let y = f(x). Then the relevant linear approx. formula is \[f(x)approxf(x _{0})+f'(x _{0})*(x-x _{0})\]
Here, f(x) = Sqrt(3+e^x) and x_0 is 0.08.
i have never used this method by i've heard of "linear approximations"
What method is your teacher (or your online instruction) using for linear approximations? Note that you are asked to approximate the value of the given function at x=0.08 using the tangent line. The formula I have given you does exactly that. \[f'(x _{0})\
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I would approximate x=0.08 to x\[\approx\]0
and solve with a point slope formula
No, Caden. You are asked to approximate the value of the given function when x=0.08. If you want to use the point-slope form of the equation of a straight line, you'll still need to determine the slope of the line; how would you do that?
i'm not sure that was just the first thought i had to answering this question
What is the name of the TOPIC you are studying right now? I call it "linear approximations." But then I realize you probably haven't studied calculus yet. What is the name of your math course?
ap calculus ab
I'm doing a practice problem and i got extremely confused so i came here
advanced placement calculus. Wow. congrats. But this means you are responsible for topics such as "linear approximations." \ If you give me a function and a particular x value (such as 0.08)
are linear approximations something i should have learned before calculus?
No, Caden, but rather "linear approximations" are an application of calculus in your first course in the subject. Perhaps your teacher is using a different name for this.
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