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EXERCISE 6.1 · Q18

Q.If f(m)=m2−3m+1f(m)=m^2-3m+1, find f(2+h)−f(2)h\dfrac{f(2+h)-f(2)}{h}, h≠0h\ne0.

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f(m)=m2−3m+1f(m)=m^2-3m+1.

f(2)=4−6+1=−1f(2)=4-6+1=-1.

f(2+h)=(2+h)2−3(2+h)+1=(4+4h+h2)−(6+3h)+1=h2+h−1f(2+h)=(2+h)^2-3(2+h)+1=(4+4h+h^2)-(6+3h)+1=h^2+h-1.

f(2+h)−f(2)=(h2+h−1)−(−1)=h2+hf(2+h)-f(2)=(h^2+h-1)-(-1)=h^2+h. …

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