High School Mathematics 2012 Final Paper Derivatives Review Solve Me

1.Knowing that the straight line y=x+1 is tangent to the curve y=ln(x+a), what is the value of a?

y'=1/(x+a)

Let the tangent point be p(s,t), and p is on the tangent y=x+1 : t=s+1 (1).

p on the curve y=ln(x+a): t=ln(s+a) (2).

Tangent slope = tangent derivative: 1 (s+a) = 1 (3).

3)==>s+a=1 substitution (2).

t=0 substituting (1) gives s=-1

a=2 2.What is the tangent equation for the curve y=x (x-2) at the point (-1,1)?

y'=-2/(x-2)²

Tangent slope k=y'|(x=-1)=-2/9

The tangent equation is: y-1=-2 9(x+1).

i.e. 2x+9y-7=0

3.If the tangent at the point p on the curve is parallel to the chord ab, what is the coordinate of the point p?

y'=4-2x, set the tangent point p(x0,y0).

ab slope kab = (4-0) (2-4) = -2

According to the title: y'|(x=x0)=4-2x0=-2

x0=3,y0=4x0-x0^2=3

p(3,3)

4.What is the monotonically increasing interval of the function f(x)=(x-3)e x?

f'(x)=e^x+(x-3)e^x=(x-2)e^x

by f'(x) >0, i.e., x-2>0 to get x>2

The monotonically increasing interval is (2,+

5.If the function f(x)=x 3+ax has two extreme points on r, what is the range of the value of the real number a?

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

f(x) has two extreme points on r.

f'(x) There are 2 different zeros.

a<06.What are the maximum and minimum values of the function f(x)=x 3-3x+1 in the closed interval [-3,0]?

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

f'(x)=0 gives x=-1, or x=1 (1 is not in the interval [-3,0], not considered).

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f(x)max=f(-1)=3 ,f(x)min=f(-)=-17

1, Derivation 2, Derivation Brought In.

The others do it themselves, and I haven't done such an easy question for many years.

1.（1）.Knowing that the maximum and minimum values of the function f(x)=1 3x3-4x+4 are a, b, respectively, try to find the value of a+b.

2) Knowing that a is a real number, the function f(x)=x3-ax2-4x obtains the extreme value at x=1, and finds the maximum and minimum values of the function on [-4,4].

Solution: (1)f'(x)=x 4 let =0 get x= 2 f (x)=2x f (2)=4 0 f ( 2)= 4 0 f(x)max=f( 2)=28 3=a, f(x)min=f(2)= 4 3=b. ∴a+b=8

2)f'(x)=2x -2ax-4 is brought in by the question f'(1)=0 to obtain a= 1, f'(x)=2x +2x-4 so that f'(x)=0 gives x=1 or x= 2 , f (x)=4x+2, f (1) = 6 0 , f (2) = 6 0, f(x)max= f( 2)=4 f(x)min=f(1)= 2.

2.Let f(x)=a x2+blnx(a, b r, and all constants) satisfy f(1)=1. Find the value of a.

Solution: Bring f(1)=1 into to obtain a=1

3.It is known that the image of f(x)=ax4+bx2+c passes through the point (0,1), and the tangent equation at x=1 is y=x-2; (1) Find the analytic formula of f(x), and (2) find the monotonically increasing interval of y=f(x).

Solution: (1) Bring (0,1) into f(x) to obtain c=1, f (x)=4ax +2bx, and know f (1)=1, 4a+2b=1 from the meaning of the questionf(1)= 1, i.e., a+b+c= 1 is obtained by a=5 2, b= 9, 2

2) f (x) = 10x -9x so that f (x) 0, 3 10 x 0, x 3 10 monotonically increasing intervals are ( 3 10 , 0) and (3,

1.(1) .Solution: f (x) = x -4

Let x -4=0 give x=-2 or 2

Combined with the derivative function image, x =-2 is the maximum point of the function, and x =2 is the minimum point of the function.

a+b=82) f (x)=3x -2ax-4 because x=1 is the extreme point of the function, a=-1 2 f (x)=3x +x-4=(3x+4)(x-1).

x=-4 3 and 1 are the two extreme points of the function (the maximum and minimum values can be obtained by comparing the two extreme values with the two endpoint function values).

2。Algebraic calculations can be found to be a=1

To add, the first step in the answer to Jiang 67219 upstairs is to derive the function f(x)=x -4

1)f'(x)=x 2+4 req f'(x)=0 The solution is x1=2 x2=-2And then the list of whatever, won't be typed with a computer. Bring x1 x2 into the calculated maximum value 28 3 minimum value -4 3 so a+b=28 3-4 3=8

The lazy one in the back is too troublesome.

I'm sorry, I haven't studied for too long, I forgot.

The slope of the tangent is y'= Since 4e baix (e x 1).

Let's consider t = e x (e x 1) = e x (e 2x 2e x 1), divide the numerator denominator du by e x, look at the zhi denominator, and use the base dao inequality to get t 1 4, so that there is y'1. In combination with the tangent function of the image, the range of inclination angles is obtained [0, 2] [3, 4, )].

No, you can only find f'(1)=2, f(1)=3, and the type of the original function f(x) cannot be determined, it may be a quadratic function, a cubic function, or some other function.

In addition, you can think about it in relation to the image, at the point (1,3) on the line y=2x+1, there can be many function images that can be tangent to the point on the line, you can draw it by hand, there must be more than one curve, so the original function cannot be determined, and its derivative function cannot be determined.

(1) Derivative of the composite function, f (x) = 2x*e (ax) + x 2 * a*e (ax) = (a*x 2+2x) e (ax).

Just discuss g(x)=a*x 2+2x, e(ax) is everbright at zero, don't think about it.

Because a<=0, when a=0, g(x)=2x, we can know that when x>=0, f(x)>=0, f(x) increases monotonically; When x<0, f(x) <0 and f(x) decreases monotonically.

When a<0, if g(x)>=0, x<=0 or x>=-2 a, then f(x)>=0, f(x) increases monotonically; When g(x) < 0, 00 is obtained, and then x>=-2 a should intersect with [0,1], that is, -2 a<=1, that is, -2=if g(x)<0, f(x) decreases monotonically, and the maximum value is f(0)=0

Find the derivative first so that the derivative = o

Then draw a frame diagram of the monotonic interval of the ball.

Remember to classify a=o a less than 0).

2) After knowing the monotonicity, find the planting of 2 inflection points, bring it into the original formula, and then bring x=1 x=0 into the original formula, compare the size of the 4 numbers to find the largest.

Let the function f(x)=ax+lnx, g(x)=a x, (1) when a=-1, find the minimum value of the distance from the point on the function y=f(x) image to the straight line x-y+3=0; (2) Is there a positive real number a that makes f(x) g(x) true for all positive real numbers x? If so, find the range of values of a?

Solution: (1) When a=-1, f(x)=-x+lnx, the slope of the straight line l:x-y+3=0 k=1.

Let f(x)=-1+1 x=1, get x=1 2, f(1 2)=-1 2+ln(1 2)=-1 2-ln2Therefore the tangent l of the point (1 2,-1 2-ln2) on the f(x) image, so the distance d from the point (1 2,-1 2-ln2) to l is minimal:

dmin=│1/2-(-1/2-ln2)+3│/√2=(4+ln2)/√2

2). ax+lnx≤a²x²，a²x²-ax-lnx≥0，a²x²-ax≥lnx

a²x²-ax=a²(x²-x/a)=a²[(x-1/2a)²-1/(4a²)]a²(x-1/2a)²-1/4≥-1/4

So only -1 4 ln(1 2a) = -ln(2a), i.e., ln(2a) 1 4,ln2+lna 1 4,lna 1 4-ln2

i.e. a e (1 4-ln2).

g(x)=?I don't understand the following formula.