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The slope of the secant line containing the two points
(x, f(x)) and (x + h, f(x + h)) on the graph of a function y = f(x) may be given as
m_sec = [f(x + h) - f(x)]/[(x + h) - x] which leads to [f(x + h) - f(x)]/h, where h cannot = 0.
Express the slope of the secant line for the function f(x) = 1/x^2 in terms of x and h. Be sure to simplify.
Let me see.
f(x) = [1/(x + h)^2 - (1/x^2)]/h
f(x) = [1/(x^2 + 2xh + h^2) - (1/x^2)]/h
After further simplifying, I got the following answer:
f(x) = (2x - h)/[(x^2)(x^2 + 2xh + h^2)]
You say?
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f(x) = (2x - h)/[(x^2)(x^2 + 2xh + h^2)]
Somewhere in that simplification you have lost a minus sign. It should be
f'(x) = (-2x - h)/[(x^2)(x^2 + 2xh + h^2)]
Then when h tends to zero it becomes f'(x) = (-2x )/[(x^2)(x^2)] = -2/x^3
Other minor point. Don't start with the secant function as f(x). f(x) is the original function not the secant function. I've used f'(x) as this is a common notation for the differential function.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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f(x) = (2x - h)/[(x^2)(x^2 + 2xh + h^2)]
Somewhere in that simplification you have lost a minus sign. It should be
f'(x) = (-2x - h)/[(x^2)(x^2 + 2xh + h^2)]
Then when h tends to zero it becomes f'(x) = (-2x )/[(x^2)(x^2)] = -2/x^3
Other minor point. Don't start with the secant function as f(x). f(x) is the original function not the secant function. I've used f'(x) as this is a common notation for the differential function.
Bob
Ok. Thanks for letting me know.
Now, does f '(x) mean f prime of x?
Is f prime of x the slope which is the derivative?
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yes.
There's two notations used but they mean the same f'(x) = dy/dx
Note. If you had z = f(x) then it would be dz/dx
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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yes.
There's two notations used but they mean the same f'(x) = dy/dx
Note. If you had z = f(x) then it would be dz/dx
Bob
I can say dy/dx reads d y d x. YES?
If I have w = f(x), I can then say dw/dx. YES?
If I had f(z), the derivative is f prime of z. Yes?
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In the UK folk usually say d y by d x . Your other statements are correct.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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In the UK folk usually say d y by d x . Your other statements are correct.
Bob
Glad to know my statements are correct.
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