Let a be greater than 0, - 1 ≤ x ≤ 1, the minimum value of function y = - x ^ 2-ax + B + 1 is - 4, and the maximum value is 0 For detailed answers, online answers read, a little do not understand

Let a be greater than 0, - 1 ≤ x ≤ 1, the minimum value of function y = - x ^ 2-ax + B + 1 is - 4, and the maximum value is 0 For detailed answers, online answers read, a little do not understand

First, when the symmetry axis X = - A / 21, it is increased in the interval - 1 ≤ x ＜ - A / 2, and decreased in the interval - A / 2 ≤ x ≤ 1
When x = - A / 2, take the maximum, because the distance from x = 1 to x = - A / 2 is much longer than that from x = - 1 to - A / 2
So when x = 1, take the minimum value, bring it in, and verify whether a meets the premise - A / 2 > 1
Hope to adopt. It's not easy to type so many words. Thank you

If y = x power of X * 2 and Y derivative equals zero, then x is?

y=x*2^x
y'=2^x+x*2^xln2=0
2^x(1+xln2)=0
Because 2 ^ X0
be
1+xln2=0
x=-1/ln2

Finding the derivative function of y equal to the x power of X

y = x^x
Take logarithm on both sides and get
lny = x * lnx
Then both sides take the derivative of X
obtain
1/y * y' = lnx + 1/x * x
So there are
y' = ylnx + y

lim(x→+∞)e^(lnx/x)-1/x^(-1/2)=lim(x→+∞)lnx/x*x^(-1/2)
I don't understand how this process is obtained, please write detailed steps, can list the formula, thank you!

When x tends to 0, e ^ X-1 and X are equivalent infinitesimals, which can be replaced by each other when calculating the limit
Here, take LNX / X as a whole. When x tends to positive infinity, LNX / X tends to 0, so e ^ (LNX / x) - 1 is equivalent to LNX / X

X tends to infinity, find the limit of LNX / X
Is "the product of bounded function and infinitesimal is infinitesimal"
Is LNX a bounded function

LNX is not a bounded function & nbsp; its range is r
Finding the limit of LNX / X
=lim  （1/x）/1=0

What is the limit of (x ^ (1 / x) - 1) ^ (1 / LNX) when x tends to infinity?

Guo Dunyong replied:
Then x ^ (1 / x) - 1) → - 1, (1 / LNX) → 0,
∴x→∞,lim[（x^(1/x) －1）^(1/lnx)] →1.

Mathematical logarithm function operation
How to calculate log 0.5 {(- x) + √ [(- x) ^ 2 + 1]} + log 0.5 [x + √ (x ^ 2 + 1)]?

Let [x + √ (x ^ 2 + 1)] = a
(-x)+√[(-x)^2+1]=B
log 0.5 A+log 0.5 B
=log 0.5 AB
therefore
Original formula = log 0.5 {(- x) + √ [(- x) ^ 2 + 1]} * [x + √ (x ^ 2 + 1)]
=log 0.5[（x^2+1）-x^2]
=log 0.5 1
=0

Ln10 reciprocal equals LGE? Why

Formula: LGA / LGB = logba (logba means base is B, index is a)
Then ln10 = LG10 / LGE = 1 / LGE

What are the ways to compare log functions? Log (7) 5 and log (6) 7; ln0.32 and LG2; what are the good ways to compare log functions?

Try to make it the same base or true number
Or compare it with numbers like 0 and 1
Here, for example, obviously log7 (5) 1
32 LG1 = 0

Find the limit: LIM (e ^ x-1-x ^ 2) / (2x + x ^ 2) when x tends to 0

Type 0 / 0
Using the law of lobida
Upper and lower derivation respectively
=lim(e^x-2x)/(2+2x)
=(1-0)/(2+0)
=1/2