If f(*)=$(1/*^2 +2* -3)d* find f(*) given that f(2) =2
If f(x)=(1x2+2x−3)dx.Find f(x) given that f(2)=2If\ f(x)=(\frac{1}{x^2} +2x -3)dx.\\ Find\ f(x)\ given\ that\ f(2) =2If f(x)=(x21+2x−3)dx.Find f(x) given that f(2)=2
This means f′(x)=x−2+2x−3f'(x)=x^{-2}+2x-3f′(x)=x−2+2x−3
Integrate f′(x)f'(x)f′(x) :
f′(x)=x−2+2x−3,∫1x2dx+∫2xdx−∫3dx=−1x+x2−3x+cf(x)=1x+x2−3x+c,But f(2)=2f(2)=2=12+22−3(2)+c,c=92 ⟹ f(x)=1x+x2−3x+92f'(x)=x^{-2}+2x-3,\intop\frac{1}{x^2}dx+\intop 2xdx-\intop 3dx=-\frac{1}{x}+x^2-3x+c\\ f(x)=\frac{1}{x}+x^2-3x+c,\\ But\ f(2)=2\\ f(2)=2=\frac{1}{2}+2^2-3(2)+c,\\ c=\frac{9}{2}\\ \implies f(x)=\frac{1}{x}+x^2-3x+\frac{9}{2}\\f′(x)=x−2+2x−3,∫x21dx+∫2xdx−∫3dx=−x1+x2−3x+cf(x)=x1+x2−3x+c,But f(2)=2f(2)=2=21+22−3(2)+c,c=29⟹f(x)=x1+x2−3x+29
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