Let y = 2/x 2 and dy =?
Solution: dy = 2 [x (-2)]' dx =-4x (-1) dx =-(4/x) dx.
2. Explain that the following functions are at the specified point.
Is it continuous? If not continuous, yes
Can you make it up?
( 1)f(x)=(x2-x-2)/(x-2); At point x=2
Solution: f (x) = (x 2-x-2)/(x-2) = (x-2) (x+1)/(x-2) = x+1.
So x=2 is its discontinuity.
2) Is f (x) = (1+xsinx)/x continuous at x=0?
Solution: Because f(0) is undefined, x→ 0lim (1+xsinx)/x.
=x→0lim[( 1/x)+sinx]=∞
∴x=0 is an infinite discontinuity of this function, and it is also an unrepairable discontinuity.
set up
y=x^(2/3)
Ask dy=
Solution: dy = (2/3) x (- 1/3) dx.
Let y = 2x 3+3x;; Find y''=?
Solution: y ′ = 6x2+3
y "= 12x
Y = ex (e-x), find y'=?
Solution: Take the logarithm of both sides and get:
lny= 1+(e-x)lnx
Take the derivative on both sides and you will get:
y '/
y=-lnx+(e-x)/x
∴y′=y[(-lnx)+(e-x)/x]=[ex^(e-x)][(-lnx)+(e-x)/x]
3. Explain whether the following piecewise functions are continuous? If it is not continuous, can it be supplemented?
( 1)f(x)= sin2x/x; (when x
f(x)= 2; (when x=0)
f(x)=(X2+2x)/x
(when x>0)
Solution: Because x→0-limf(x)=x→0-lim(sin2x/x).
=
x→0-lim(2x/x)=2
x→0+limf(x)= 1
x→0+lim(x^2+2x)/x
=
x→0+lim(x+2)=2
∴x→0limf(x)=f(0)=2
So the function is continuous at x=0.
y=ln( 1+x2)
, find y'=?
Solution: y ′ = 2x/(1+x 2)
Y = e (2x), find y''=?
Solution: y ′ = 2e (2x)
y″=4e^(2x).
Point out the discontinuity of the following functions and explain whether they can be supplemented.
f(x)=sin3x/x
Solution: ∫x→0 limb(x)= 1
x→0 LIMS in 3 x/x = x→0 LIMS in 3 x/x = 3
∴x=3
Is the complementary discontinuity of this function.
f(x)=2x- 1
(When x≤ 1
f(x)=(X2
–1)/(x-1) (when x >; 1)
Solution: when x> is at 1, f (x) = (x 2-1)/(x-1) = x+1.
So x→ 1+limf(x)= 1
x→ 1+lim(x+ 1)=2
And f (1) = 2-1=1,x→ 1-LIMF(x)= 1
x→ 1-lim(2x- 1)= 1
So x→ 1 limb(x) does not exist, that is, x= 1 is an irreducible discontinuity of this function.