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????
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
Well, let's take a look at this problem and see if we can find the right answer.
When finding the local linear approximation of a function at a specific point, we use the equation:
L(x) = f(a) + f'(a)(x - a)
Where f(a) is the value of the function at the point of interest, f'(a) is the derivative of the function at that point, and (x - a) represents the distance from the point of interest to the point where we want to find the approximation.
In this case, we want to find the local linear approximation of f(x) = e^(3x) at x = 1.
Let's start by finding f(1), which is simply plugging x = 1 into the function:
f(1) = e^(3*1) = e^3
So, we have f(1) = e^3.
Next, we need to find f'(1), the derivative of the function at x = 1.
Taking the derivative of f(x) = e^(3x) with respect to x, we get:
f'(x) = 3e^(3x)
Plugging x = 1, we have f'(1) = 3e^3.
Now that we have f(1) = e^3 and f'(1) = 3e^3, we can write the equation for the local linear approximation:
L(x) = e^3 + 3e^3(x - 1)
So, the correct answer is c) y = 3e^(3)(x − 1).
To find the local linear approximation of f(x) = e^(3x) at x = 1, we can use the formula for linear approximation.
The formula for linear approximation is:
L(x) = f(a) + f'(a)(x - a)
where f'(a) is the derivative of f(x) evaluated at x = a.
In this case, a = 1, so we need to find f(1) and f'(1).
f(1) = e^(3(1)) = e^3
To find f'(x), we take the derivative of f(x) with respect to x:
f'(x) = d/dx(e^(3x)) = 3e^(3x)
Now, we can evaluate f'(1):
f'(1) = 3e^(3(1)) = 3e^3
Substituting these values into the formula for linear approximation, we get:
L(x) = e^3 + 3e^3(x - 1)
So, the correct answer is c) y = 3e^(3)(x − 1).
To find the local linear approximation of a function at a particular point, we can use Taylor's theorem, which states that any differentiable function can be approximated by a polynomial of certain degree.
In this case, we want to find the local linear approximation of the function f(x) = e^(3x) at x = 1. The linear approximation is a first-degree polynomial in the form y = mx + b, where m is the slope and b is the y-intercept.
To find the slope, we need to find the derivative of f(x) and evaluate it at x = 1. Taking the derivative of f(x) = e^(3x) with respect to x, we get:
f'(x) = 3e^(3x)
Evaluating f'(x) at x = 1, we have:
f'(1) = 3e^(3*1) = 3e^3
This gives us the slope of the linear approximation.
To find the y-intercept, we substitute x = 1 into the original function f(x) = e^(3x):
f(1) = e^(3*1) = e^3
Therefore, the linear approximation is y = 3e^3(x - 1) + e^3, since the slope is 3e^3 and the y-intercept is e^3.
Comparing this with the provided options:
a) y = e^3: This is not the correct linear approximation.
b) y = e^(3(x - 1)): This is the correct linear approximation.
c) y = 3e^(3)(x - 1): This is not the correct linear approximation, as the slope is missing the factor of 3 and the y-intercept is incorrect.
d) y = 3e^(3)x - 2e^3: This is not the correct linear approximation, as the y-intercept is incorrect.
Therefore, the correct answer is option b) y = e^(3(x - 1)).