When the value of independent variable x is, the value of function y = 3x-15 is less than 0? When the value of independent variable x is, the value of function y = 2X-4 is greater than 0=

When the value of independent variable x is, the value of function y = 3x-15 is less than 0? When the value of independent variable x is, the value of function y = 2X-4 is greater than 0=


x<5 x≥2



Function y = {- x + 1, x > 0; 0, x = 0; X + 1, X


Step 1: judge the size of X and 0 step 2: if x is greater than 0, then y = - x + 1 step 3: if x = 0, then y = 0 step 4: if x is less than 0, then y = x + 1 by the way, what's your optimization plan?



If y = [f (x ^ 2)] ^ (1 / x), where f (x) is a differentiable positive function, how to find dy!


Take logarithm:
lny=1/x lnf(x^2)
Derivation of X: y '/ y = - 1 / x ^ 2 LNF (x ^ 2) + 1 / X * f' (x ^ 2) * 2x
y'=y[-1/x^2 lnf(x^2)+2f'(x^2)]
The results are as follows
dy=y[-1/x^2 lnf(x^2)+2f'(x^2)]dx