Questions about derivatives, questions about derivatives

1) Let a(x1,f(x1)) b(x2,f(x2)), then.

k(ab)=[f(x1)-f(x2)]÷x1-x2)[x1^3+ax^2+b-(x2^3+ax2^2+b)]÷x1-x2)

x1^3-x2^3+ax1^2-ax2^2)÷(x1-x2)[(x1-x2)(x1^2+x1x2+x2^2)+a(x1-x2)(x1+x2)]÷x1-x2)

x1^2+x1x2+x2^2+a(x1+x2)x1^2+(a+x2)x1+x2^2+ax2＜1x1^2+(a+x2)x1+x2^2+ax2-1＜0△1=(a+x2)^2-4*(x2^2+ax2-1＞0-3x2^2-2ax2+a^2+4＞0

2=(-2a)^2-4*(-3)(a^2+4)＞04a^2+12＞0 ∴a∈r

2)f'(x)=3x^2+2ax

k=3x^2+2ax(0＜x＜1)

k|≤1|3x^2+2ax|1 -1 3x 2+2ax 1 is constant on x (0,1).

a [-root number 3, -1].

The above steps are reversible.

k|The sufficient condition for 1 is a [-root number 3, -1].

The first question upstairs was wrong.

As you can see from the title, there must be a tangent line at one point on the image that is parallel to the straight line of the line connecting any two points, that is, the slope is equal.

Derivative function f'(x)=-3x^2+2ax<1

Then the problem is transformed into a problem where the parabolic vertices are located below the x-axis, and the solution of a 2<12 is obtained by a (-root 3, root 3).

y=lnx-4/x³-3/x^4

y=f(x)=c

c is a constant), then f'(x)=0

f(x)=x^n

n is not equal to 0).

f'(x)=nx^(n-1)

x n denotes x to the nth power of x).

f(x)=sinx

f'(x)=cosx

f(x)=cosx

f'(x)=-sinx

f(x)=a^x

f'(x) = a xlna(a>0 and a is not equal to 1, x>0) f(x) = e x

f'(x)=e^x

f(x)=logax

f'(x)=1/xlna

a>0 and a is not equal to 1, x >0).

f(x)=lnx

f'(x)=1/x

x>0)

f(x)=tanx

f'(x)=1/cos^2

xf(x)=cotx

f'(x)=-

1/sin^2

The method of operation of the x-derivative is as follows.

f(x)+/g(x))'f'(x)+/

g'Filial piety is grinding (x).

f(x)g(x))'f'(x)g(x)+f(x)g'Marking bridges (x) g (x) f(x)).'f(x)'g(x)-g(x)f'(x))/f(x))^2

Here are all the derivative formulas.

Take a good look.

Volume: v = sh = hr 2 = 1 (constraint).

Material dosage: min: m = 2 r 2 + 2 rh = 2 h + 2 h (objective function).

DM DH = -2 H + H = 0 (extreme condition).

h = (4) height of the drinking tank).

r = 1 ( h) = h2 base radius).

Let the radius of the bottom surface be r and the height be h

Given that pie r 2h=1, find the minimum value of 2 pie r 2+2 pie rh h=1 (pier 2), and substitute it to obtain.

2 pie r 2 + 2 r

Let f(x)=2 x 3+2 x, then f'(x)=4 faction x-2 (x 2) when x=3 times root number (1 (2 faction)), f'(x)=0, so when the radius of the bottom surface is 3 times under the root number (1 (2 pies)) and the height is 3 times under the root number (4 pies), the material used is the most economical.

When the ratio of the height to the bottom radius of the metal beverage can is 2:1, the material used can be minimized.

1.Derivation, is a quadratic equation, and finding the minimum value of this quadratic function is x02You got the second derivative wrong.

The derivatives of the two high stoves are of the same sign, and the extreme points must be checked whether the derivatives on both sides are different signs.

4.Synopsis, let discriminant = 0

5.No, fractional inequalities must not be done in this way, they must be divided first.

6.If it's not a big problem, you can draw a picture to test it, if it's a big problem, set a tangent point (x0, y0), export the tangent expression, let it pass (2, Qing Nian brother 2), and finally solve 3 x0.

7.One of the tangents of the second function crosses the origin, so let's make sense, ditto, and derive the tangent expression.

8.y=(1+ x) 2+1-f(2)= x 2+2 x divide.

(1) Because g(x) is continuous on r.

So g(x) is continuous at x=0 point.

That is, lim(x->0)g(x)=g(0).

lim(x->0)f(x)/x=a

Because f(x) is continuous in the second derivative on r, and f(0) = 0

So according to Lobida's law, lim(x->0)f'(x)=a

a=f'(0)

2) Since g(x) is continuous on r, g(x) is continuous on r, and from the conclusion of the above question, the determined value of a is f'(0)

Because when x≠0, g(x) = f(x) x,g'(x)=[xf'(x)-f(x)] x 2, apparently g'(x) Continuous on x≠0.

Now prove that when a=f'(0), g'(x) is continuous at x=0 points.

g'(0)=lim(t->0) [g(t)-g(0)]/t

lim(t->0) [f(t)/t-f'(0)]/t

lim(t->0) [f(t)-tf'(0)]/t^2

lim(t->0) [f'(t)-f'(0)]/2t

f''(0)/2

Because lim(x->0) g'(x)=lim(x->0) [xf'(x)-f(x)]/x^2

lim(x->0) [f'(x)+xf''(x)-f'(x)]/2x

lim(x->0) f''(x)/2

f''(0)/2

g'(0) So when a=f'(0), g'(x) is continuous at x=0 points.

That is, g(x) is continuous in the derivative of the previous order r.

1,f'(x)=ax^2-3x+a+1

f'(1)=0

a=12,f'(x) > Cheong Tsai x 2-x-a+1 is a suspicion. x^2+2)a-2x>0

a>2x Grip (x 2+2).

So. 2x/(x^2+2)≤0x≤0

Solution: Because y'=6x-4, so the slope of the tangent of the curve through the (1,1) point is k=6*1-4=2

The straight line is parallel to this tangent, so the slope of the straight line is 2The straight line passes through the point p(-1,2), so the equation for the straight line is y-2=2(x+1), that is, y=2x+4

f'Hail volt (x) = x 2-2x-3

f'(x)=0

x^2-2x-3=0

x-3)(x+1)=0

x1=3 x2=-1

y great source with ants = y|x=-1 =1/3-1-3+3=2/3y|Minimum-y|x=3 =9-9-9+3=-6