Find the equation of the line tangent to
f(x) = xe^-x
at the point where x = 0. What does this tell you about the behavior of the graph when x = 0?
first of all , when x = 0 , f(0) = 0
so the point is (0,0)
f'(x) = x(-e^-x) + e^-x
when x = 0
f'(0) = 0 + 1 = 1
equation of tangent : y = x
at (0,0) the function is increasing
To find the equation of the line tangent to the graph of the function f(x) at the point where x = 0, we need to find the derivative of the function and evaluate it at x = 0.
Step 1: Find the derivative of f(x)
The function f(x) = xe^-x is a product of two functions: x and e^-x. We can use the product rule to find the derivative.
Using the product rule:
f'(x) = (x * e^-x)' = x' * e^-x + x * (e^-x)'
= 1 * e^-x + x * (-e^-x)
= e^-x - xe^-x
Step 2: Evaluate the derivative at x = 0
To find the equation of the tangent line at x = 0, we need to evaluate f'(x) at x = 0.
f'(0) = e^0 - 0 * e^0
= 1 - 0
= 1
Step 3: Use the point-slope form to write the equation of the tangent line
The equation of a line in point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line, and m is the slope of the line.
Since the tangent line passes through the point (0, f(0)), and we now have the slope m = f'(0), we can write the equation of the tangent line as:
y - f(0) = f'(0)(x - 0)
Plugging in the values, we get:
y - 0 = 1(x - 0)
y = x
So, the equation of the tangent line to the graph of f(x) = xe^-x at the point where x = 0 is y = x.
This tells us that the graph of the function has the same slope as this tangent line at x = 0. In other words, the graph is "flattening out" or becoming less steep at this point.