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how do I find the slope of the curve at x = -9. when Let f(x)=5 x^2 + 2/x^2. (this new to me)
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I first find the f'(x) = 10x+1/-2x right ?
or dy/dx= 10x-4x^-3 right
then I plug in for x=-9
yes correct
then it is 10(-9)-4(-9)^3=
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-90-2916=
oh the exponent is neg, so you are dividing by 9^3
wait is ot 10(-9)-4(-9)^-3
that is correct --> -90 +4/9^3
(-9)^-3 = -1/9^3
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wait im lost
so its -90-1/9^3
you did something wrong when you wrote ...-90 -2916 property of neg exponents: x^-n = 1/x^n right
yes
10(-9) - 4(-9)^-3 = -90 +4/9^3 = -90 +4/729 = -89.9945
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oh ok divide 4/729
I was multiplying 4(9)^3
:)
is that the approximate or exact answer?
how can i find the exact number because they want me to find the exact anwser
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nevermind I found it thanx
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