Please answer all Please answer all Give an example of a polynomial P(x) of degr

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Please answer all Give an example of a polynomial P(x) of degree 5 for which x = 1 is a critical number but not a local minimum, nor a local maximum. (Hint: An example of a polynomial of degree k is P(x) = (x-a)^k (2--5) Determine the intervals on which the function is decreasing and increasing and then find local minima and maxima. f(x) = (x-2)(x+3) 2) f(x) = (x+1)(x-2)(x+3) 3) f(x) = x e^(-x) 4) f(x) = x^x defined on the interval (0, infinity). 5) Let f(x) = cos x+2 cos^2 x defined for pi/2

Explanation / Answer

1) Let P(x) = (x-1)5

P(x) is of degree 5

And P'(x) = 5(x-1)4 = 0 => x=1 so 1 is critical point of P(x)

P"(x) = 20(x-1)3 =0 at x=1 and so x=1 is neither a point of minimum nor a point of maximum

2) f(x) = (x+1)(x-2)(x+3)

f'(x) = 3x2+4x-5 = 0 => x=-2.1196, 0.78630

So we look for intervals (-infinity, -2.1196), (-2.1196, 0.78630), (0.78630,infinity)

On (-infinity, -2.1196), f'(x) >0 and so the function is increasing on (-infinity, --2.1196)

On (-2.1196, 0.78630), f'(x)<0 and so the function is decreasing on (-2.1196,0.78630)

On (0.78630, infinity), f'(x) >0, so the function is increasing on (0.78630)

3) f(x) = xe-x

f'(x) = e-x-xe-x = e-x(1-x) = 0 => x=1

We will consider (-infinity,1) and (1,infinity)

On (-infinity,1), f'(x)>0, so the function is increasing on (-infinity,1)

On (1,infinity), f'(x)<0, so the function is decreasing on (1,infinity)

4) f(x) = xx

f'(x) = xx(logx+1) = 0 => x=1/e = 0.36788

We will consider (0,0.36788), (0.36788,infnity) because the function is defined on (0,infinity)

Now on (0,0.36788), f'(x) <0 and so the function is decreasing on (0,0.36788)

On (0.36788, infinity), f'(x) >0 and so the function is increasing on (0.36788,infinity)

5) f(x) = cosx+2cos2x

f'(x) = -sinx-4cosxsinx = -sinx(1+4cosx) =0 => x=1.82348

Consider (pi/2, 1.82348) and (1.82348,pi)

On (pi/2, 1.82348), f'(x) <0 and on (1.82348,pi), f'(x) >0

Hence x=1.82348 is point of minimum

Also f"(x) = -cosx-4cos2x = 0 => x=2.45335

Hence x= 2.45335 is inflection point of f(x)

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