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a) f(x)=ln(x), so f'(x)=1/x. By MVT, f'(c) = (f(b)-f(a))/(b-a) = (ln b - ln a)/(b-a) = ln(b/a)/(b-a), where c lies between a and b.  Now, since 1/x is a decreasing function, and a < c < b, we get 1/...

Taking a sample survey is quicker, cheaper and easier to carry out than a census. 

4-[(x+3)*((x-2)/(x^2 +5x +6))]

4-[(x+3)(x-2)/(x^2 +5x +6))]

factorise denominator 

4-[(x+3)(x-2)/(x+3)(x+2)]

cancel down (x+3)

4-[(x-2)/(x+2)]

expand 

4(x+...

dy/dx = 3x^2-2x+6

At (1,5), dy/dx = 7. 7 is the gradient of the tangent, therefore the gradient of the normal at (1,5) is -1/7.