High School Research Mathematics Topic The Application of Derivatives in Life 20

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From the equation e x(x+a+1)=0 at least one real root on (0, positive infinity), first find the root of the equation, because e x cannot be equal to 0, so x+a+1=0, so x=-a-1, it can be seen that the equation has only one real root is x=-a-1, because this real root is on (0, positive infinity), so -a-1>0, a<-1 can be solved.

Analytical blNA>ALNB

lna/a>lnb/b

f(x)=lnx x is proved in (e,+ is the subtractive function (1-lnx) x <0,x>e.

ln[(a to the power b) divided by (b to the power a)]=ln[(a to the power of a)-ln(to the power of b) = b*ln(a)-a*ln(b)>b*ln(a)-a*ln(a)>0; So, ln[(a to the power b) divided by (b to the power of b)] > ln1; So [(a to the power b) divided by (b to the power of b)] > 1

So, the power of b of a and the power of a of b.

s'（t）=3t²-6t

Since t is time, t 0

When a(3+root31) 2, s(a) s(a+1) when (3+root31) 2 a, s(a) s(a+1) when (3+root31) 2=a, s(a)=s(a+1).

Substituting t t into the equation separately to get s respectively, and then dividing by the corresponding t can be obtained.

s(t) derivative s'(t)=3t 2-6t=3(t-1) 2-3 is a function of velocity and time It can be seen that the velocity is greater than or equal to t=a when t=a+1.

is continuous on r, so f is known to be'(x)=3ax 2+1 has two zeros, discriminant =-12a 0, so the range of a is (negative infinity, 0)f'(x) The set of 0 solutions is (-root-1 3a, root-1-3a), f'(x) The 0 solution set is (negative infinity, -root-1 3a), (root-1 3a, positive infinity), so the monotonically increasing interval is (-root-1 3a, root-1 3a) and the monotonically decreasing interval is (negative infinity, -root-1 3a), (root-1 3a, positive infinity).

2。Constructor.

g(x)=x-x^2/2-ln（1+x）

The derivative yields g'(x)=1- x-1 1+x=-x 2 1+xWhen x>0, g'(x) 0 is constant, so g(x) decreases monotonically on (0, positive infinity).

So when x>0, g(x) g(0)=0

Hence the limit of x-x 2 20.

Question 1: First, discuss whether a is equal to 0 If it is 0, then f(x)=x is obviously not in line with the topic. Therefore, a is not 0.

Finding the derivative of f(x) is equal to 3ax 2+1 Obviously, a>0 is not true, because it is a single increase, which is not in line with the topic. Therefore, a<0, i.e., a, the range of values of a is a<0, and the monotonic interval is based on the two values that are solved so that the derivative is equal to 0 in the case of a<0 are the poles.

Question 2: Let f(x)=x-x 2 2-ln(1+x) Since they are all elementary functions, then f(x) must be derivable Finding the derivative gives f'(x)=1-x-1 (1+x) =-x 2 (1+x)< 0 is constant then f(x) "f(0) is also constant and f(0)=0 so f(x)=x-x 2 2-ln(1+x)" 0 is true i.e.

X-X 2 2 Hope it helps!

The derivative function is f'(x)= 2x +2axx=0 x=4. The derivative is 0, and the derivative is substituted. 32+ 8a=0 a=-4

Substituting the original function. x=0. y=b x=4, y=64-16+b=48+b, so b is the minimum value -1

a=-4 b=-1

Solution: Derived from the question.

f'(x)=3x^2+2ax.i.e. when x=0 and x=4 are equations f'(x)=0. So f'(4)=48+8a=0, i.e. a=-4;

And because the minimum is -1, by f'(x). At x>4,x<0, f(x) increases monotonically. When 0 is then taken at x=4 to the minimum. i.e. f(4)=4 3-4*4 2+b=-1. So b=-1.

a=-1;b=-1