Draw the image of function y = X-2 and judge its monotonicity

Draw the image of function y = X-2 and judge its monotonicity

First, draw the image of y = x, which is a straight line passing through the origin with a slope of 1, and then translate it down two units, which is the image of y = X-2. Obviously, it increases monotonically

Draw the graph of function f (x) = 3x + 2, judge its monotonicity and prove it

Let x = 0, y = 2. Let y = 0, x = - 23, and the image of function f (x) = 3x + 2 is a straight line passing through point (0, 2), (- 23, 0), as shown in the figure. From the image of function, we know that function f (x) = 3x + 2 is an increasing function. It is proved as follows: let x1, X2 be taken at any position (- ∞, + ∞), let X1 < X2, f (x1) - f (x2) = (3x1 + 2) - (3x2 + 2) = 3 (x1-x2) < 0, and function f (x) = 3x + 2 It's an increasing function

This paper discusses the domain of definition, range of value, parity and monotonicity of the function y = x ^ (2 / 5), and draws the image of the function

y=x^(2/5)=（5）√（x^2)
Domain R
Range y ≥ 0
Even function
X ≥ 0, single increment
x

The tangent equation of the function f (x) = ax + blnx + C (a, B, C are constants) at x = e is (E-1) x + ey-e = 0, and f (1) = 0. (I) find the values of constants a, B, C; (II) if the function g (x) = x2 + MF (x) (m ∈ R) is not a monotone function in the interval (1, 3), find the value range of real number M

Let f (x) be (0, + ∞), f ′ (x) = a + BX, the tangent equation of F (x) at x = e is (E-1) x + ey-e = 0, f ′ (E) = − e − 1E, and f (E) = 2-e, that is, a + be = − e − 1E, AE + B + C = 2-e, and f (1) = a + C = 0, the solution is a = - 1, B = 1, C = 1 (5 points) (II) from (I), we know that f (x) = - x + LNX + 1 (x ＞ 0) ∧ g (x) = x2 + MF (x) = x2 MX + mlnx + m (x ＞ 0) ∧ g ′ (x) = 2x − m + MX = 1x (2x2 − MX + m) (x ＞ 0) ∧ (7) let D (x) = 2x2 MX + m (x > 0). (I) when G (x) has an extreme value in (1,3), G '(x) = 0 has and only has one root in (1,3), that is, D (x) = 2x2 MX + M = 0 has and only has one root in (1,3), and ∵ D (1) = 2 > 0, when d (3) = 0, that is, M = 9, D (x) = 2x2 MX + M = 0 has and only one root in (1,3), x = 32, and（ 3) When G (x) has two extremum in (1,3), G '(x) = 0 has two roots in (1,3), that is, the quadratic function d (x) = 2x2 MX + M = 0 has two unequal roots in (1,3), so △ = M2 − 4 × 2 × M > 0d (1) = 2 − m + m > 0d (3) = 2 × 32 − 3M + m > 01 < M4 < 3 In conclusion, the value range of real number m is (8, + ∞) (13 points)

Let f (x) = ax square + 2x + blnx get the extremum when x = 1 and x = 2, 1: find the analytic expression of the function, 2: find the function in【

Y '= 2aX + 2 + B / x = 0, two are 1,22a + 2 + B = 0,4a + 2 + B / 2 = 0, the solution is a = - 1 / 3, B = - 4 / 3

The extremum of the function y = ㏑ x-x & # 178; is?

The domain is (0, + infinity)
The derivative is y '= 1 / x-2x = 0
X = 1 / radical 2
x> 1 / radical 2, y '

It is known that x = 1 is an extreme point of the function f (x) = AlN (1 + x) + x ^ 2-10x
It is known that x = 1 is an extreme point of the function f (x) = AlN (1 + x) + x ^ 2-10x
And the monotone interval of function f (x)

f(x)=aln(1+x)+x^2-10x
f'(x)=a/(1+x))+2x-10
X = 1 is an extreme point
f'(1)=a/2+2-10=0
a=16.
Domain x > - 1,
f'(x)=16/(1+x)+2x-10=(16+2x^2+2x-10x-10)/(1+x)
=(2x^2-8x+6)/(1+x)
=2(x^2-4x+3)/(1+x)
F '(x) > 0, - 1 < X3, increasing interval (- 1,1), (3, + ∞)
f'(x)

Let f (x) = X3 + 3bx2 + 3CX at two extreme points x1, X2, and x1 ∈ [- 1,0], X2 ∈ [1,2]
Then the sum of the maximum and minimum of F (x2) is
Can we eliminate the 3C solution? C belongs to [- 1,0] f (x2) belongs to [- 16-2c, - 2-3c] and the result is - 15
See the second question of jingyou.com for answers
b→c a→b
3b=-3x2^2-6ax2
f(x2)=x2^3+3ax2^2-3x2^3-6ax2^2
=-2x2^3-3ax2^2
f‘（x2）=-6x2^2-6ax2

Today, I have nothing to tell you: don't ask questions without thinking. If you want to ask for advice from others, you should prepare all the relevant questions! Half of the questions, no links, no pictures! Except me, who can see your broken questions? The questions are not clear! I don't know what you are asking! How can I answer you? Can I eliminate the 3C solution? What are you asking? F

The function f (x) = x & # 178; - 2lnx, G (x) = X-2 √ x.1 is known. It is proved that f (x) = g (x) + 2 has a unique solution when x > 0

The value range of function f (x) - [g (x) + 2] = (X & # 178; - X-2 + 2 √ x) - 2lnx is investigated;
The definition domains of the above functions are: (0, + ∞);
When x → + 0, Lim {f (x) - [g (x) + 2]} = Lim {(X & # 178; - X-2 + 2 √ x) - 2lnx} = Lim {- 2-2lnx} = + ∞;
When x → + ∞, Lim {f (x) - [g (x) + 2]} = Lim {(X & # 178; - X-2 + 2 √ x) - 2lnx} = Lim {X & # 178;} = + ∞;
Therefore, the function under investigation has a minimum in its domain of definition;
Let {f (x) - [g (x) + 2]} '= 2x-1 + (1 / √ x) - (2 / x) = 0 → → 2x + 1 / √ x = 2 / x + 1, we can see that x = 1 is a solution of the left equation;
The right end function (2 / x + 1) of the equation 2x + 1 / √ x = 2 / x + 1 is a simple decreasing function when x > 0, while the left end function (2x + 1 / √ x) has a minimum value of 3 / 2 ^ (1 / 3) (corresponding to x = x0 = 1 / 2 ^ (1 / 3)). Therefore, when x > x0, the equation may have (if any, only one) solution, that is, x = 1 is the unique solution;
When 0

The zeros of the function f (x) = x ^ 3-4x are

—2；0；2