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Mathematics 21 Online
OpenStudy (anonymous):

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)

OpenStudy (anonymous):

I first find the f'(x) = 10x+1/-2x right ?

OpenStudy (anonymous):

or dy/dx= 10x-4x^-3 right

OpenStudy (anonymous):

then I plug in for x=-9

OpenStudy (dumbcow):

yes correct

OpenStudy (anonymous):

then it is 10(-9)-4(-9)^3=

OpenStudy (anonymous):

-90-2916=

OpenStudy (dumbcow):

oh the exponent is neg, so you are dividing by 9^3

OpenStudy (anonymous):

wait is ot 10(-9)-4(-9)^-3

OpenStudy (dumbcow):

that is correct --> -90 +4/9^3

OpenStudy (dumbcow):

(-9)^-3 = -1/9^3

OpenStudy (anonymous):

wait im lost

OpenStudy (anonymous):

so its -90-1/9^3

OpenStudy (dumbcow):

you did something wrong when you wrote ...-90 -2916 property of neg exponents: x^-n = 1/x^n right

OpenStudy (anonymous):

yes

OpenStudy (dumbcow):

10(-9) - 4(-9)^-3 = -90 +4/9^3 = -90 +4/729 = -89.9945

OpenStudy (anonymous):

oh ok divide 4/729

OpenStudy (anonymous):

I was multiplying 4(9)^3

OpenStudy (dumbcow):

:)

OpenStudy (anonymous):

is that the approximate or exact answer?

OpenStudy (anonymous):

how can i find the exact number because they want me to find the exact anwser

OpenStudy (anonymous):

nevermind I found it thanx

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