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Let the function f (x) be a derivable even function with a period of 5 on R, then the slope of the tangent of the curve y = f (x) at x = 5 is () A. −1 five B. 0 C. 1 five D. 5
∵ f (x) is a derivable even function on R,
The image of F (x) is symmetrical about the Y axis,
The extreme value of F (x) is obtained at x = 0, that is, f '(0) = 0,
And the period of ∵ f (x) is 5,
‡ f ′ (5) = 0, that is, the slope of the tangent of curve y = f (x) at x = 5 is 0,
Therefore, the option is B
Let the function f (x) be a derivable even function with a period of 5 on R. then the tangent slope of the curve y = f (x) at x = 5?
The slope of the tangent at x = 5 is
k=f'(5)=lim(∆x->0)[f(5+∆x)-f(5)]/(∆x)
=lim(∆x->0)[f(∆x)-f(0)]/(∆x)
K = f '(5) = LIM (∆ X - > 0) [f (5) - f (5 - ∆ x)] / (∆ x)
=lim(∆x->0)[f(0)-f(-∆x)]/(∆x)
=lim(∆x->0)[f(0)-f(∆x)]/(∆x)
=-lim(∆x->0)[f(∆x)-f(0)]/(∆x)
=-k
So: k = 0
Note: the above uses that the derivative function approximates and derives from the left, and approximates and derives from the right. The two are equal
Let the function f (x) be a derivable even function with a period of 5 on R, then the slope of the tangent of the curve y = f (x) at x = 5 is () A. −1 five B. 0 C. 1 five D. 5
∵ f (x) is a derivable even function on R,
The image of F (x) is symmetrical about the Y axis,
The extreme value of F (x) is obtained at x = 0, that is, f '(0) = 0,
And the period of ∵ f (x) is 5,
‡ f ′ (5) = 0, that is, the slope of the tangent of curve y = f (x) at x = 5 is 0,
Therefore, the option is B
Let the function f (x) be a derivable even function with a period of 5 on R, then the slope of the tangent of the curve y = f (x) at x = 2.5 is?
f(x)=f(x+5)
f'(x)=f'(x+5)
f(x)=f(-x)
Then f '(x) = - f' (- x)
That is, but the function is odd
So f '(2.5) = - f' (- 2.5)
So f '(2.5) = - f' (- 2.5 + 5)
f'(2.5)=0
So the slope is 0
Let function f (x) be a derivable even function on R and satisfy f (x-3 / 2) = - f (x + 5 / 2), then the tangent slope of curve y = f (x) at x = 8 is
If f (x) is an even function, the derivative of F (x) at x = 0 is 0
F (x-3 / 2) = - f (x + 5 / 2) is transformed into f (x + 5 / 2) = - f (x-3 / 2)
f(8)=f(11/2+5/2)=-f(11/2-3/2)=-f(8/2)=-f(3/2+5/2)=f(0)
So the derivative of F (8) is 0
Therefore, the tangent slope is 0
It is known that the tangent slope of the function f (x) = x ^ 3-6ax + 8 at x = 1 is - 3 (1) Find the value of real number a; (2) Find the monotone interval of function f (x); (3) Find the maximum and minimum of function f (x) on interval [- 2,2] Online waiting for answers! How many questions will you write? Thank you~~
(1) To find the derivative first, f (x) derivative = 3x ^ 2-6a
Because the tangent slope of the function f (x) = x ^ 3-6ax + 8 at x = 1 is - 3
So f (1) = - 3, 3-6a = - 3, a = 1
(2) F (x) derivative > 0,3x ^ 2-6 > 0, a radical 2
F (x) lead
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