High School Math Derivatives Problems No Points, Help .

Your first question doesn't seem to be right: a=1 b=-2a=-2 f(x)=x 2-2x+2 The second condition can find that the axis of symmetry is x=1 (it should be).

The second asked: g'(x)=x 2 * e x - ex 2 When x=1, the derivative is zero and happens to be the minimum point, the minimum is 2e 3 (just discuss it with an image), the maximum is not (when x is less than 1, the derivative is always less than zero, the upstairs seems to be a bit of a problem, and the minimum seems to be miscalculated).

I'm not here for the distribution, I wish you success in the college entrance examination.

Are you sure your parsed formula for f(x) is correct?

1. f(1+x)=f(1-x) means that the axis of symmetry is x=1, which can be written as f(x)=a(x-1) 2+2-a through a(3,5) point, then a=1, that is, f(x)=(x-1) 2+1=x 2-2x+2

2、g'(x)=f'(x)e^x+f(x)*(e^x)'-e/3(x^3)'

2x-2)e^x+(x^2-2x+2)*e^x-ex^2x^2(e^x-e)

Ream'(x)=0 gives x=0, x=1

g when x>1'(x)>0

When x<0 g'(x)>0

When 0 so x=0 takes the maximum value g(0) = f(0) = 2, so at x = 0 takes the minimum value g(1) = f(1) e-e 3 = 5e 3

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f(x)=lnx-1/4x+3/4x -1 g(x)=x^2-2bx+4

If any x1 belongs to (0,2), and there is x2 that belongs to [1,2], and f(x1) is greater than or equal to g(x2), then the value of the real number b is obtained.

Solution: For any f(x1)>=g(x2), f(x1) is the maximum value and g(x2) is the minimum value.

f(x)=lnx-1 4x+3 4x -1 =f(x)=lnx+1 2x -1 in monotonically increasing (0,2), f(x1)=f(2)=ln2

g(x)=x 2-2bx+4 x range[1,2], categorical discussion x=-(b 2a) x=b

When a=1 4, it is a subtraction function on f(x)(0,1) and an increasing function on (1,2).

So for any 0 i.e. 2bx x +9 2, i.e. 2b x + 9x 2, in the range of [11 2, 17 4].

Therefore, 2b 11 2, the solution gives b 11 4, that is, the range of the real number b is [11 4,+

Your problem doesn't seem to be quite right, right: 1 4x+3 4x, this can be combined, let me provide an idea, it should be to find the minimum gas of function 1 in the range (use the derivative to find it) in finding the maximum value of function 2 in the range to ensure that function 1 is greater than function 2 is a very simple problem.

Straighten out your thoughts:

Condition for existence of minima: f'(x) = 0 and f''(x) >0f'(x) = 0 ==> 3x^2-3b = 0 ==> b = x^2

f''(x) >0 ==> 6x > 0 ==> x > 0 because the value of x is on (0,1) and the value of b is also on (0,1).

It is best to learn the solution ideas of this topic from the top students.

- ,0) is the increase-leakage nanoband function, and (-0) is f'(x)=6x^2-6(a+1)x+6a>0

x^2-(a+1)x+6>0,(-0)

Upper function image, single return to reed tone, you

1. (a+1)/2>0

2.(A+1) 2 "Ruler = 0, δ< 0

Condition 1 gives f(1).'=1.

The derivative is defined as the limit of f(x+c)-f(x) c when c is approaching zero. f(1-x)-f(1+x)/3x

Write (f(1-x)-f(1)+f(1)-f(1+x)) 3x=(f(1-x)-f(1)) 3x+(f(1)-f(1+x)) 3x=-f(1)).'/3+(-f(1)')3=-2/3

Choose B. Ha.

Give you the answer and the process of solving the problem.

Solution: Derivative y'=x 2, p point is on the curve, so the slope of the tangent is 2 2=4, so the tangent equation is y=4(x-2)+4=4x-4.

If a given point (a,b) is no longer on the curve, let the tangent of the point be tangent to the curve and the point (t,1 3t 3+4 3), then the slope of the tangent is: (t-a), on the other hand, the tangent of the curve at the point (t,1 3t 3+4 3) is t 2, so there is:

t-a)=t 2, and then solve the equation to find t.

Here p is the tangent point.

y'=x²x=2,k=y'=4

So y-4=4(x-2).

4x-y-4=0

If not, assume b(p,q).

The tangent point is (a,a 3+4 3).

y'=x then the slope is a

y-(a³/3+4/3)=a²(x-a)

After bq-(a 3+4 3)=a (p-a), a can be solved.

If p is not a tangent point, then the tangent has another tangent point.

Find the tangent equation for the curve. Substitute the p-spot in.