"The line x+y = 0 is tangent to the graph of y = F(x). Find F(x) if F'(x) = x".
I know how to find the antiderivative of F'(x), but I don't know how to use the tangent line to find my constant.
so the line x+y gives a slope of -1
f(x)=1/2 x^2+c but at some x, the tangent line has a slope of -1
f'= x so x=-1
what is y when x=-1? y=1
so the point of tangency is -1,1
f(x)=1/2 x^2 + C
1=1/2+C
C= 1/2
check my thinking.
Thank you! That makes sense!
To find F(x), we need to integrate F'(x) = x. The antiderivative of x with respect to x is 1/2 * x^2 + C, where C is a constant of integration.
To find the value of the constant C, we need to use the given information that the line x + y = 0 is tangent to the graph of y = F(x).
First, let's rewrite the equation x + y = 0 in terms of y:
y = -x
Since the line is tangent to the graph of y = F(x), at the point of tangency, the slopes of the line and the graph of F(x) are equal.
The slope of the line x + y = 0 can be found by rearranging the equation in slope-intercept form (y = mx + b):
y = -x + 0
Comparing this to the slope-intercept form, we can see that the slope, m, is -1.
The slope of the graph of F(x) at any point is given by F'(x). Therefore, we know that F'(x) = -1 at the point of tangency.
We are given that F'(x) = x, so to find the point of tangency, we set F'(x) = -1 equal to x:
x = -1
Substituting x = -1 into the equation y = F(x), we can find the corresponding y-coordinate:
y = F(-1)
Now, let's integrate F'(x) = x:
F(x) = 1/2 * x^2 + C
Since the point of tangency is (-1, y), we can substitute x = -1 and y = F(-1) into the equation:
F(-1) = 1/2 * (-1)^2 + C
F(-1) = 1/2 + C
We know that the line is tangent to the graph at this point, so the slope of the line is equal to the slope of the graph of F(x) at this point.
The slope of the line x + y = 0 is -1. Comparing this to the slope of the graph of F(x), we have:
F'(-1) = -1
We can differentiate the equation F(x) = 1/2 * x^2 + C to find F'(-1):
F'(-1) = 1 * (-1) + 0
F'(-1) = -1
Setting F'(-1) = -1 equal to F'(-1) = -1, we can solve for the constant C:
-1 = -1/2 + C
Adding 1/2 to both sides:
-1 + 1/2 = C
Simplifying:
-1/2 = C
So, the constant C is -1/2. Therefore, the function F(x) is:
F(x) = 1/2 * x^2 - 1/2
To find F(x), we need to integrate F'(x) = x. Since F'(x) represents the derivative of F(x), we can integrate the right-hand side of the equation to find F(x):
∫x dx = 1/2x^2 + C
Here, the constant of integration (C) represents the unknown constant term in the antiderivative. To find the value of C, we need to use the given information that the line x + y = 0 is tangent to the graph of y = F(x).
Since the line is tangent to the graph, it means that the slope of the line is equal to the derivative of F(x) at the point of tangency. In other words, F'(x) is equal to the slope of the tangent line, which is the negative reciprocal of the coefficient of x in the equation of the line, since the line is of the form x + y = 0.
In this case, the coefficient of x in the equation x + y = 0 is 1. So, the slope of the tangent line is -1.
Therefore, setting F'(x) = -1, we have:
x = -1
Now we can substitute this into the antiderivative we found earlier:
1/2(-1)^2 + C = -1
Simplifying, we get:
1/2 + C = -1
To isolate C, we subtract 1/2 from both sides:
C = -1 - 1/2
C = -3/2
Now we can substitute this value of C back into the antiderivative we found earlier:
F(x) = 1/2x^2 - 3/2
Therefore, F(x) = 1/2x^2 - 3/2 is the function that satisfies the given conditions.