so the curve is given as well as the point where the tangent touches the curve. The equation for the curve is y= 1/(x+2) and the tangent line touches at x=3. I have one answer but the book has a different answer. I did it like 3 times and i still have the same answer. So i just want to check which answer is right. thanks
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Taylor Swift is Awesome:
y = 1/(x+2)
Using quotient rule:
dy/dx = -1/(x+2)²
at x = 3
dy/dx = -1/(3+2)² = -1/25 <--- This is the slope of the tangent line.
y = 1/(x+2)
Using quotient rule:
dy/dx = -1/(x+2)²
at x = 3
dy/dx = -1/(3+2)² = -1/25 <--- This is the slope of the tangent line.
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Hi.
The answer I got was -1/25. If that is not what the book has then don't read the rest; I apologize for wasting your time. But if the book does have that answer, then please read below to see how I solved.
y= 1/(x+2)
We can think of this also as ---> (x+2)^-1 because the negative exponent makes the numbers switch from numerator to denominator.
Ok now we need to find the equation of the derivative. You do this by multiplying a variable by its exponent, then subtracting 1 from the existing exponent to become the new exponent. Kind of hard to describe in words....
SO with (x+2)^-1... We multiply (x+2) by -1 since -1 is our exponent.
[] -(x+2)^( -1)
Next we subtract 1 from the existing exponent. -1-1 is -2.
[]-(x+2)^( -2) becomes our equation for the derivative.
Now with the derivative equation, we plug in 3 for x to find out the slope of the tangent line at x=3.
Plug it in----> the slope is (-1/25)
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CLIFF NOTES
{}Convert your original equation to the equation of the derivative in 2 steps
[1'] Multiply the variable by its exponent
[2] subtract 1 from the exponent to get the new exponent.
{}Plug in your x value into the derivative equation to find the slope of the tangent line at that x value.
The answer I got was -1/25. If that is not what the book has then don't read the rest; I apologize for wasting your time. But if the book does have that answer, then please read below to see how I solved.
y= 1/(x+2)
We can think of this also as ---> (x+2)^-1 because the negative exponent makes the numbers switch from numerator to denominator.
Ok now we need to find the equation of the derivative. You do this by multiplying a variable by its exponent, then subtracting 1 from the existing exponent to become the new exponent. Kind of hard to describe in words....
SO with (x+2)^-1... We multiply (x+2) by -1 since -1 is our exponent.
[] -(x+2)^( -1)
Next we subtract 1 from the existing exponent. -1-1 is -2.
[]-(x+2)^( -2) becomes our equation for the derivative.
Now with the derivative equation, we plug in 3 for x to find out the slope of the tangent line at x=3.
Plug it in----> the slope is (-1/25)
=================================
CLIFF NOTES
{}Convert your original equation to the equation of the derivative in 2 steps
[1'] Multiply the variable by its exponent
[2] subtract 1 from the exponent to get the new exponent.
{}Plug in your x value into the derivative equation to find the slope of the tangent line at that x value.