if x + 7y = 29 is an equation of the line normal to the graph f at the point (1,4), the f '(1) =
To find the gradient of the graph f at the point (1,4), we need to find the derivative of the function f(x) and evaluate it at x = 1.
Since we are given the equation x + 7y = 29, we can rearrange it to solve for y:
x + 7y = 29
7y = 29 - x
y = (29 - x)/7
Now, let's find the derivative of f(x):
f'(x) = d/dx [(29 - x)/7]
= -1/7
To find f'(1), we substitute x = 1 into the derivative:
f'(1) = -1/7
Therefore, f'(1) = -1/7.
x + 7y = 29 has a slope of -1/7
Since it is normal to the graph at (1,4), f'(1) = 7
To find the slope of the line normal to the graph of f at the point (1,4), we need to determine the slope of the tangent line to the graph of f at that point and then find the negative reciprocal of that slope.
First, let's rewrite the equation x + 7y = 29 in slope-intercept form (y = mx + b), where m is the slope of the line:
x + 7y = 29
7y = -x + 29
y = (-1/7)x + 29/7
The slope of the tangent line to the graph of f at (1,4) is the coefficient of x in the equation, which is -1/7.
The slope of the line normal to the graph of f at (1,4) is the negative reciprocal of -1/7. To find it, we take the reciprocal (flipping the fraction) and then change the sign:
-1/7 reciprocal = -7/1 = -7
Therefore, f'(1) = -7.
Apologies for the mistake in my previous response.
If the equation x + 7y = 29 represents a line normal to the graph f at the point (1,4), then the slope of the line would be the negative reciprocal of the derivative of f at x = 1.
The equation x + 7y = 29 can be rearranged to solve for y:
7y = 29 - x
y = (29 - x)/7
This equation represents the function f(x). To find f'(x), we take the derivative of f(x) with respect to x:
f'(x) = d/dx [(29 - x)/7]
= -1/7
To find f'(1), we substitute x = 1 into the derivative:
f'(1) = -1/7
Therefore, f'(1) = -1/7, not 7.