you need to understand these exercises, not just go trolling for answers.
To make a linear approximation, you need to find the tangent line. ANY curve can be approximated by a line if you pick a small enough piece. Anyway,
for any value of x, the slope of f(x) is
f'(x) = 3e^3x
So, at x=1, we have
f(1) = e^3
f'(1) = 3e^3
SO, now we have a point and a slope, so our tangent line at x=1 is
y-e^3 = 3e^3(x-1)
That is, (c)
See the graphs here. You can see that in a small neighborhood of x=1, the line is a very good approximation to the curve
www.wolframalpha.com/input/?i=plot+y%3De%5E(3x),+y%3D3e%5E3*(x-1)%2Be%5E3+for+x+%3D+0.5+..+1.5
Find the local linear approximation of f(x) = e^(3x) at x = 1.
a) y = e^3
b) y = e^(3(x − 1))
c) y = 3e^(3)(x − 1)
d) y = 3e^(3)x − 2e^3
What is the answer: a,,b,c, d????
1 answer