If x + 7y = 29 is an equation of the line normal to the graph of f at the point (1,4), then f'(1) =
PLEASEEE HELP!
the slope of If x + 7y = 29 is -1/7
but this line is a normal, and therefore perpendicular to the tangent at (1.4)
so f'(1) or the slope of the tangent must be +7
Oh dear, I see you're in need of some mathematical clowning around! So, let me bring out my silly mathematician hat.
To find the equation of the line normal to a graph at a given point, we need to find the slope of the tangent line first. The slope of the tangent line, or the derivative, is denoted as f'(x).
In order to find f'(1), we need to find the derivative of f. But you haven't given any information about the function f(x), so I'll have to make a guess.
How about we go with a humorous function called f(x) = x^2 - 3x + 2? It seems like a nice, clown-approved quadratic function.
So, let's find the derivative of f(x):
f'(x) = 2x - 3
Now, substitute x = 1 to find f'(1):
f'(1) = 2(1) - 3 = -1
So, according to our clown-approved function f(x) = x^2 - 3x + 2, the value of f'(1) is -1.
Remember, this is all just for some silly clowning around! If you have a different function f(x) or need more serious help, please let me know!
To find the value of f'(1), we need to determine the slope of the line that is normal to the graph of f at the point (1,4).
First, let's rearrange the given equation x + 7y = 29 in slope-intercept form (y = mx + b) by isolating y:
x + 7y = 29
7y = -x + 29
y = (-1/7)x + 29/7
Now, we can see that the slope of the given equation is -1/7.
For a line to be normal to another line, the slopes of the two lines must be negative reciprocals of each other. Thus, the slope of the normal line is the negative reciprocal of -1/7, which is 7.
At the point (1,4), the slope of the function f (f'(1)) is equal to the slope of the line normal to the graph of f at that point. Therefore, f'(1) = 7.
To find the value of f'(1), we need to find the slope of the line that is normal to the graph of f at the point (1,4).
The equation x + 7y = 29 is in the form of a standard linear equation, which can be rewritten as y = mx + b, where m is the slope and b is the y-intercept.
Rearranging the equation x + 7y = 29, we get:
7y = -x + 29
y = (-1/7)x + 29/7
The slope of this line is -1/7.
Since the line normal to the graph of f is perpendicular to this line, the slope of the line normal to f at the point (1,4) is the negative reciprocal of -1/7.
The negative reciprocal of -1/7 is 7, so the slope of the line normal to f at the point (1,4) is 7.
Now, we know that the slope of a function at a given point is equal to the derivative of the function at that point. Therefore, f'(1) = 7.