Let f(x) = (x + 1 )x. None of the rules that we\'ve learned seem to help us find
ID: 2982271 • Letter: L
Question
Let f(x) = (x + 1 )x. None of the rules that we've learned seem to help us find f'(x), but we will find that's not true! Explain why neither the power rule nor a formula like d/dx 3x = 3xln3 will help you to compute f'(x). Start with the equation y = (x + 1)x, take the natural logarithm of both sides, and write out the equation t=you get. Use your previous knowledge to find the derivative of the right-hand side, ln((x + 1)x). Now use implicit differentiation to find the derivative d/dx ln y. Use the fact that ln y = ln ((x + 1 )X) to solve for the derivative f'(x) = dy/dx in terms of x alone (you may have to substitute).Explanation / Answer
f'(x) = 0 => 4x^3 -12x^2 =0
critical points : x=0,0,3
f''(x) =12x^2 - 24x
f''(x) >0 for x=3. Hence 3 is a local minima.
=0 for x=0
f'''(x) =24x-24 < 0 for x=0
=> x=0 is neither local maxima nor local minima. It is an inflection point.
So, only x=3 is local minima and there are no local maxima
b) f(x) >0 for x = -inf
f(x) decreases from x=-inf to x=3 and increases from 3 to +inf
c) x=0 is the inflection point.
d)f(x) is concave up if f''(x)>0 i.e x>2 or x<0
concave down in o<x<2
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