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anonymous
 one year ago
A. Find the derivative of f at x. That is, find f'(x).
B. Find the slope of the tangent line to the graph of f at each of the two values of x given to the right of the function
f(x)=x^28; x=1, x=3
anonymous
 one year ago
A. Find the derivative of f at x. That is, find f'(x). B. Find the slope of the tangent line to the graph of f at each of the two values of x given to the right of the function f(x)=x^28; x=1, x=3

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.2you know the derivative of \(x^2\)?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2if the answer is "no" that is fine, i will tell you

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2and the derivative of a constant is zero, making the derivative of \(f(x)=x^28\) the function \[f'(x)=2x\] that is all

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but its two parts to the question is 2x the answer to them both

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2no \(2x\) is the derivative you now need the equation for two lines, one where \(x=1\) and the other where \(x=3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f(x)=1^28 and f(x)=3^28

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2if \(x=1\) then \(y=(1)^28=7\) so the point is \((1,7)\) and the slope is \(2\times 1=2\) use the point slope formula to get the equation of the line

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2if \(x=3\) then \(y=3^28=1\) the point is \((3,1)\) and the slope is \(2\times 3=6\) points slope again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y(7)=2(x(1)) and y(1)=6(x(3))

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2you can clean them up a great deal, but yes
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