The graph of x-2y=9 is parallel to the normal line to the graph of f(x) when x=5. What is the value of f'(5)? Justify your answer.
x-2y=9 slope is 1/2 slope that belongs to the normal line is -2 not sure what else to do
ok I am reading this incorrectly. Maybe my slope is 1/2
slope of \(x-2y=9\) is \(\frac{1}{2}\) so slope of normal line is \(-2\)
oh i see you said that there is nothing else to do except say \(f'(5)=-2\) that is all
Note: the slopes of two lines that are perpendicular are negative reciprocals, which satellite73's comment aptly illustrates.
You say you're in precalculus, but I see you finding the derivative: f'(x). How's that?
parallel to the normal line?
My username is precal but I tend to study calculus for fun
Calculus is not my strong point, I struggle with many calculus concepts. Solving 600 pages of calculus just for fun.
It surely adds to your problem solving skills if you understand that the derivative of a function represents the slope of the tangent line to the curve of that function at a certain point / certain x-value.
Let's go back to the beginning. Re-read the original question. What are you sure you are able to do correctly? What are you not so sure about?
The graph of x-2y=9 is parallel to the normal line. y=(-9/2)+(1/2)x is the function and the slope is 1/2 that is parallel to the normal line of f(x). I am assuming that f(x) is some other function and not x-2y=9
What leads you to assume that f(x)=-9/2 + (1/2)x?
Let's suppose that we DON'T know f(x). What is the slope of the TANGENT line to the graph of this function at x?
No, I am not assuming that f(x) is that. I solved for y to determine the slope. I don't know anything about f(x) except that this line is parallel to the normal line
m=-2 this slope is the slope of the tangent line
because the tangent line and normal line are perpendicular
All right. Let's back up a bit. You are dealing with a function named f(x), right? What is the derivative of f(x)?
This is some type of AP Calculus AB problem, I know I am overlooking something.
f'(x)
that's right. Now, what is the slope of any line that is PERPENDICULAR to this tangent line? Write it symbolically.
-1/f'(x)
Yes, my point exactly.
Now, what is the slope of that given line?
1/2
yes, both lines are parallel and therefore have the same slope
Sorry, my sentence was 'way off. I'm going to delete it after re-writing my question.
We are told that the given line and the normal line to the curve at x=5 are parallel. What does that say ab out their slopes?
so is my answer -1/-2 or 1/2
so satellite was incorrect? Not even sure I am correct but I am following your logic
other slope is negative reciprocal
now I am not sure, is the solution -2 or 1/2
slope is 1/2 in the beginning
slope of x-2y=9 is 1/2
I agree that the slope of the given line is 1/2. You correctly found an expression for the slope of the normal line. For review, type it out again here. Now equate those two quantities and solve for f'(5).
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