A mathematical problem related to the second function of junior high school Well, it's actually very simple. The image of a linear function y = - x + 2 doesn't pass through any quadrant The answer is the third one, I know. But if the value of X is negative, then the image will pass through the fourth quadrant

A mathematical problem related to the second function of junior high school Well, it's actually very simple. The image of a linear function y = - x + 2 doesn't pass through any quadrant The answer is the third one, I know. But if the value of X is negative, then the image will pass through the fourth quadrant

When x is negative and Y is positive, how can it be the third quadrant after passing through the second quadrant?
If you want to pass through the third quadrant, X and y need to be negative, but if x is negative,
Y must be positive because y = 2-x, x < 2
I don't know how to ask,

If the inverse function of the exponential function y = FX passes through the image point (2, - 1), then the exponential function is?

Substituting (2. - 1) into
-1=2*F F=-1/2
Y=-1/2X

Given the function f (x) = 2Sin (Π - x) cos x 1. Find the minimum positive period of F (x). 2. Find the maximum and minimum of F (x) in the interval [- Π / 6, Π / 2]

F (x) = 2Sin (Π - x) cos x = 2sinxcosx = sin2x, so the minimum positive period is π
The maximum value of F (x) in the interval [- Π / 6, Π / 2] is f (π / 4) = 1, and the minimum value is f (- π / 6) = - √ 3 / 2

F (x) = 2cos & sup2; α + √ 3sin2 α + B, when x belongs to 0 to π / 2, f (x) has the maximum value?

F (x) = 2cos & sup2; α - 1 + √ 3sin2 α + B + 1

Trigonometric function
Given 2tana = 3tanb, prove Tan (a-b) = sin2b / 5-cos2b

tan(A-B)=(tanA-tanB)/(1+tanAtanB)=tanB/(2+3tanBtanB)
=sinBcosB/(2cosBcosB+3sinBsinB)
=sin2B/(6-2cos2Bcos2B)
=sin2B/(5-cos2B)

Trigonometric number of compulsory four
Help me to prove: 1-2sinxcosx / cos ^ 2-sin ^ 2 = 1-tanx / 1 + TaNx
sin^4x+cos^4x=1-2sin^2xcos^2x
Thank you. You'd better give me a hint and let me do it myself

1. It's (1-2sinxcosx) / (COS ^ 2x-sin2 ^ x) = (1-tanx) / (1 + TaNx)!
Left molecule (1-2sinxcosx) = (COS ^ x-sin ^ x) ^ 2
Left denominator (COS ^ 2x-sin2 ^ x) = (COS ^ x-sin ^ x) (COS ^ x + sin ^ x)
Then left = (COS ^ x-sin ^ x) / (COS ^ x + sin ^ x)
The right cut chord is also (COS ^ x-sin ^ x) / (COS ^ x + sin ^ x)
So: (1-2sinxcosx) / (COS ^ 2x-sin2 ^ x) = (1-tanx) / (1 + TaNx)
two
1-2sin^2xcos^2x=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=sin^4x+cos^4x

Mathematics problem (concept of function 1.2.1)
Given that G (x) = 1-2x, f [g (x)] = 1-x Λ 2 / X Λ 2 (x ≠ 0), then f (1 / 2) is equal to?

g(1/4)=1-1/2=1/2
f[1/2]=f[g(1/4)]=[(1-(1/4)^2]/(1/4)^2=15

Suppose that the profit function of an enterprise is l (q) = 10 + 2q-0.1q & # 178;, what is the output Q with the maximum profit?

L（q）=10+2q-0.1q²=-0.1（q²-20q-100）=-0.1（q-10）²+20
It can be seen from the above formula that when (Q-10) = 0, l (q) is the largest, that is, when the profit is the largest, the output is 10 and the profit is 20

Given the profit L (q) = 10 + 2q-0.1q ^ 2, how many units of production is the maximum profit? And the maximum profit can be obtained

Q is output, right?
The profit is a quadratic function of Q. obviously, the maximum value of the quadratic function is what the subject asks for
When q = - B / 2A = - 2 / (2 * (- 0.1)) = 10, the profit is the largest
Just bring in the equation, l = 20

It is known that the total revenue function of an enterprise is r = 26q-2q ^ 2-4q ^ 3, and the total cost function is C = 8q + Q ^ 2
It is known that the total revenue function of an enterprise is r = 26q-2q ^ 2-4q ^ 3, and the total cost function is C = 8q + Q ^ 2, where q is the output of the product,
Seeking the output and maximum profit when the enterprise obtains the maximum profit

A:
The total revenue function is r = 26q-2q ^ 2-4q ^ 3, and the total cost function is C = 8q + Q ^ 2
Then the profit function:
y=R-C=26Q-2Q^2-4Q^3-8Q-Q^2=-4Q^3-3Q^2+18Q
The derivation of Q is as follows
y'(Q)=-12Q^2-6Q+18
=-6*(2Q^2+Q-3)
=-6(2Q+3)(Q-1)
When - 3 / 2