I think linear approximation? (square root, tangent, e^x)

[meredith]

Homework Statement

the value of f(x) = (sqrrt e^x +3) at x=0.08 obtained from the tangent to the graph at x=0 is...?

Homework Equations

The Attempt at a Solution

i used linear approximation.
(sqrrt e^o +3) + (1/2(sqrrt3+e^0)(0.08)
i got an answer but i know its wrong. i got like 1.72 or something.
did i do it all wrong?

 

[D H]

Your use of parentheses doesn't make sense. Do you mean

f(x) = \sqrt{e^x+3}

f(x) = \sqrt{e^x} + 3

f(x) = \left(\sqrt e\right)^{\;x + 3}

f(x) =\left(\sqrt e\right)^{\;x} + 3

Or even yet something else?

 

[meredith]

Your use of parentheses doesn't make sense. Do you mean

f(x) = \sqrt{e^x+3}

f(x) = \sqrt{e^x} + 3

f(x) = {\sqrt e}^{x + 3}

f(x) = {\sqrt e}^x + 3

Or even yet something else?

yes i meant the first one sorry i don't know how to do that stuff!

 

[D H]

You could have written it as f(x)=sqrt(e^x+3) and that would have been fine.
f(x)=(sqrt e^x+3) was pretty much meaningless.

What is the derivative of f(x) at x=0?

 

[meredith]

You could have written it as f(x)=sqrt(e^x+3) and that would have been fine.
f(x)=(sqrt e^x+3) was pretty much meaningless.

What is the derivative of f(x) at x=0?

would that be 1/2sqrrt(1+3) = 1/4?
so then i multiply that by 0.08
and add it to 2.
ok i got it thanks!