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Find the approximately value of f(5.001), …

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Find the approximately value of f(5.001), where f(x) =x²-7x²+15.

    • Posted by Pavan Dubey 6 years, 1 month ago

    CBSE > Class 12 > Mathematics
    • 2 answers

    Christine Phoebe 6 years, 1 month ago

    Let x=5 then x+d x =5.001 dx=0.001 y=x^2 - 7x^2 + 15 = - 6x^2 +15 Differentiating dy= (- 12x+0) dx dy= - 12 × 0.001 = - 0.012 Now f(5)= -6(5×5)+15= - 135 Answer would be f(5)+ dy Which is - 135.012

    0Thank You

    Khuskaran Sidhu 6 years, 1 month ago

    Put x=5.001

    0Thank You

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