If point (2,8) is on the image of power function y = f (x), then f (- 1) + F (1)=

If point (2,8) is on the image of power function y = f (x), then f (- 1) + F (1)=

Power function y = x ^ n
Substituting (2,8)
8=2^n
n=3
f(x)=x³
f（-1）+f（1）=-1+1=0

Find the inverse function of y = 2 ^ X. (x power of 2)

From y = 2 ^ x we get x = log2 (y),
Exchange X and y to get y = log2 (x), which is the inverse function

Given that the inverse function of the function f (x) = 3 passes through point (18, a + 2), let g (x) = 3 of ax power - 4 of x power be defined as interval [- 1,1]
(1) Find the analytic expression of G (x) (2) if the equation f (x) = m has a solution, find the real number range of M
(3) For any n ∈ R, try to discuss the number of solutions of the equation g (| x |) + 2 | x | + 1 power = n

1) The inverse function of F (x) passes through the point (18, a + 2) so that f (x) passes through the point (a + 2,18), that is, 3 ^ (a + 2) = 18, then 3 ^ a = 2, so g (x) = 3 ^ (ax) - 4 ^ x = (3 ^ a) ^ x-4 ^ x = 2 ^ x-4 ^ (x) 2) f (x) = m has a solution, that is, to find the range of F (x) and the definition field of X is r, so f (x) = 3 ^ (x) = m > 03) Let f (x) = g (| x | + 2 ^ (...)

How to find the inverse function of negative x power with y equal to 3

Let AB not equal to 0 and b > a find the coordinates of the intersection point of the image of the first-order function y = ax + b y = BX + a

y=ax+b=bx+a
(a-b)x=(a-b)
b> A then A-B ≠ 0
So x = 1
y=a+b
So the intersection (1, a + b)

If the image of the function y = ax + B passes through one, three and four quadrants, then the function y = BX + A does not pass through the third quadrant

The graph of function y = ax + B passes through one three four quadrants,
Then the line intersects the x-axis on the positive half axis and the y-axis on the negative half axis
Let x = 0, then y = B ｜ B0 ｜ a > 0
The function y = BX + a intersects the X axis at (- A / B, 0)
Intersection with y axis at (0, a)
∵-a/b>0,a>0
The function y = BX + A does not pass through the third quadrant

Let f (x), G (x) be differentiable, and find the derivatives of the following functions: 1) y = 1 + F & sup2 under the root sign; (x) + G & sup2; (x) 2) F & sup2 of y = E; (x) the power of X × f (E)

F (x), G (x) can be derived everywhere
1) Y = f (x + e to - x power)
y' = f'[x+e^(-x))]*[1-e^(-x)]
2) Y = f (x power of E) × g (x) power of e
y' = e^x * f'(e^x)*e^g(x) + f(e^x)*e^g(x) * g'(x)
3)y={xf(x²）}²
y' = 2xf(x^2)*[f(x^2) + x*2x * f '(x^2)]

How to find the derivative of e ^ - X - 1 (the - x power of e minus 1) to have an inverse function? Otherwise, how to determine whether a function has an inverse function

How to find the derivative of e ^ - X - 1 (E's - x power minus 1): [e ^ (- x) - 1] '= [e ^ (- x)]' = e ^ (- x) * (- x) '= - e ^ (- x). Does a differentiable number have an inverse function? Otherwise, how to determine whether a function has an inverse function: whether there is an inverse function has nothing to do with whether it is differentiable

① How to find the derivative of the eighth power of y = (5x - 4)? How to find the derivative of 2x - 5 under y = radical?

The first question I think = 40 (5x-4) ^ 7
y=u^8 u=(5x-4)
Derivation of dy / Du = (u ^ 8) = 8U ^ 7
The derivation of Du / DX = (5x-4) = 5
dy/dx=(dy/du)*(du*dy)=8u^7*5=40u^7
Bring u = (5x-4) in
=40（5x-4）^7
This is my understanding
I'm the same as downstairs

Find the derivative of the following function at a given point (1) y = x to the power of 1 / 4, x = 16 (2) y = SiNx, x = Pie / 2 (3) y = cosx, x = pie
Process, online, etc

(1)
y=x^0.25
y'
=0.25x^(0.25-1)
=0.25x^(-0.75)
y'(16)=0.25/8=1/32
(2)
y=sinx
y'=cosx
y'(π/2)=0
(3)
y=cosx
y'=-sinx
y'(2π)=-1