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If the function y = f (x) is differentiable in the interval (a, b) and x0 € (a, b), then lim f(x0+h)-f(x0-h)/h h->0 The value of is?
lim [f(x0+h)-f(x0-h)]/h
= 2 lim [f(x0+h)-f(x0-h)]/2h
= 2 f'(x0)
Application proof of higher derivative Let the function f (x) be continuous on [0, a], differentiable in (0, a), and f (0) = 0, f '(x) monotonically increases, so that G (x) = f (x) / X. it is proved that G (x) is an increasing function There seems to be a mistake on the first floor
Construct f (x) = f (x) x to derive this function, it is easy to know its monotonic increase, and then use the definition method to calculate g (x1) - G (x2) = f (x1) x1-f (x2) x2 / x1x2. Finally, it is easy to get that G (x) is an increasing function!
Two trucks are driving on two vertical highways and close to each other. One is driving from west to east at a speed of 90km / h, and the other is driving from north to South at a speed of 60km / h. When the first truck is 200m away from the highway intersection and the second truck is 150m away from the intersection, what is the speed at which the two trucks approach each other?
S from the meaning of the question ²= (s a) ²+ (s b) ² (where s is the distance between two trucks)
The derivative of T on both sides is 2S (DS / DT) = 2 (s a) (DS A / DT) + 2 (s b) (DS B / DT)
Substitute s = (200 ²+ one hundred and fifty ²)^ (1 / 2) = 250 s a = 200 s b = 150
Ds a / dt = 90 DS B / dt = 60
dS/dt=108
That is, the speed at which the two trucks approach each other is 108km / h
Application of derivative in Higher Mathematics It is proved that the equation 4x = 2 ^ X has and has only one real root in (0,1) The process of this problem is to prove that the function has at least one real root in the (0,1) interval by using the zero value theorem Then the monotonicity of the function is judged by finding its first derivative. It is proved that the first derivative of the function is greater than 0 It shows that it is a monotonically increasing function. Finally, it is proved that the function has in the (0,1) interval And there is only one real root I don't understand two points of this problem 1. Why can we prove that the function is monotonically increasing, and then we can get that the function has and only one is in the (0,1) interval root 2. If it is proved that the function is monotonically decreasing, can it still show that there is and only one real root!
1. Let f (x) = 4x-2 ^ x, it has been proved that the function increases monotonically and has a root. Let the root be a (then f (a) = 0), then a belongs to (0,1)
Then, when x is f (x) < 0 on (0, a) and f (x) > 0 on (a, 1), there is only f (a) = 0
2. Yes. As long as it is a monotonic function. The proof is the same as above, but at this time, when x is on (0, a), f (x) > 0; F (x) < 0 on (a, 1)
Ask: a proof of derivative application in Higher Mathematics Prove that the area of the triangle does not exceed [(3 times the root sign 3) multiplied by (the square of R)] / 4, where R is the radius of the circumscribed circle. (Application of derivative)
prove:
Known from sine theorem
A / Sina = B / SINB = C / sinc = 2R, R is the circumscribed circle radius of triangle ABC
The area of triangle ABC can be expressed as s = (1 / 2) * a * b * sinc, where C is the included angle of a and B
Substitute a = Sina * 2R, B = Sina * 2R
S=(1/2)*sinA*2R*sinB*2R*sinC=sinAsinBsinC*2R ²
To prove s
Advanced Mathematics (derivative) Let the function f (x) = 1 / x, find f ^, (- 1) Can you write it in detail?
F "(x) = - x ^ - 2 substitute (- 1) into - 1
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