Derivative: y = (INX) / (x + 1) - 3 ^ x

Derivative: y = (INX) / (x + 1) - 3 ^ x

y'=[1/x*(x+1)-lnx*1]/(x+1) ²- 3^x*ln3
=(x+1-xlnx)/[x(x+1) ²]- 3^x*ln3

Function f (x) = the square of X * the derivative of F (2) - 3x, find the derivative of F (3). It's so difficult, ask for help ~ No points, sorry, heroes

f(x)＝x^2*f'(2)－3x
Derivative f '(x) = 2xf' (2) - 3 (because f '(2) is a constant)
Take x = 2
f'(2)＝2*2*f'(2)－3
Get f '(2) = 1
Bring back the original formula and you can get f (3) = 3

If the derivative of function f (x) is known to be f '(x) and f (x) = 3x2 + 2xf' (1), then f '(3) = () A. 9 B. 6 C. -6 D. 20

f′（x）=6x+2f′（1），
Let x = 1, get f ′ (1) = 6 + 2F ′ (1), and solve f ′ (1) = - 6,
Then f '(x) = 6x-12,
So f '(3) = 6 × 3-12=6，
Therefore, B

Given that f (x) = (m-1) x squared + 3x + (2-N), and this function is an odd function, find the values of M and n

F (x) = (m-1) x squared + 3x + (2-N) is an odd function on R
Then there must be f (0) = 0
So n = 2
f(-x) = (m-1)(-x)^2+3(-x)=(m-1)x^2-3x=-f(x)
2(m-1)x^2=0
Because x is not equal to 0
So m = 1
So m = 1, n = 2

Given the function f (x) = 3x square - 5x + 2, find the values of F (negative radical 2), f (- a), f (a + 3), f (a) + F (3)

Substitute the value in or not? F (negative root 2) = 3 * 2 + 5 times root 2 + 2 = 5 + 5 times root 2
f(-a)=3a^2+5a+2
f(a+3)=3(a+3)^2-5(a+3)+2=3a^2+13a-4
f(a)+f(3)=3a^2-5a+2+27-15+2=3a^2-5a+16

Given x > 0, how to find the minimum value of the function f (x) = x square - 3x + 1 / x

It is solved by mean inequality
f(x)=(x ²- 3x+1)/x
=x-3+1/x
=(x+1/x)-3
≥2√[x(1/x)]-3
=2-3
=-1
If and only if x = 1 / x, i.e. x = 1, the equal sign holds
The minimum value is - 1