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Find delta f and df when f(x) = 2x^2 - 1, x = 1, delta x = 0 cdot 02

Answer»

SOLUTION : `f(x) = 2x^2 - 1, x = 1, delta x = 0 cdot 02`
Then `delta f = f(x+ Delta x) - f(x)`
=f(1.02) - f(1)
= `2 (1 cdot 02)^2 - 1 - (2- 1)`
=`2 cdot 0808 - 2 = 0 cdot 0808`
Again `df= 4X DX = 4 xx 1 xx 0 cdot 02 = 0 cdot 08`


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