1.
a)
f ( x ) = ln ( − 2 x 2 − x − 6 ) + x 2 − 1 f(x)=\ln(-2x^2-x-6)+\sqrt{x^2-1} f ( x ) = ln ( − 2 x 2 − x − 6 ) + x 2 − 1
− 2 x 2 − x − 6 > 0 -2x^2-x-6>0 − 2 x 2 − x − 6 > 0 x 2 − 1 ≥ 0 x^2-1\ge0 x 2 − 1 ≥ 0
D = ( − 1 ) 2 − 4 ( − 2 ) ( − 6 ) = − 47 < 0 D=(-1)^2-4(-2)(-6)=-47<0 D = ( − 1 ) 2 − 4 ( − 2 ) ( − 6 ) = − 47 < 0 Then − 2 x 2 − x − 6 < 0 , x ∈ R -2x^2-x-6<0, x\in \R − 2 x 2 − x − 6 < 0 , x ∈ R
D o m a i n : { } Domain:\{\} Do main : { }
b)
f ( x ) = ln ( − 2 x 2 − x − 6 ) + x 2 − 1 f(x)=\ln(-\dfrac{2}{x^2-x-6})+\sqrt{x^2-1} f ( x ) = ln ( − x 2 − x − 6 2 ) + x 2 − 1
x 2 − x − 6 < 0 x^2-x-6<0 x 2 − x − 6 < 0 x 2 − 1 ≥ 0 x^2-1\ge0 x 2 − 1 ≥ 0
( x + 2 ) ( x − 3 ) < 0 (x+2)(x-3)<0 ( x + 2 ) ( x − 3 ) < 0
x ≤ − 1 o r x ≥ 1 x\le -1\ or\ x\ge1 x ≤ − 1 or x ≥ 1 Domain:( − 2 , − 1 ] ∪ [ 1 , 3 ) (-2, -1]\cup[1,3) ( − 2 , − 1 ] ∪ [ 1 , 3 )
2.
𝑓 ( 𝑥 ) = x 2 − 4 x + 5 , x ∈ [ 3 , 4 ] 𝑓(𝑥)=x^2−4x+5,x\in [3, 4] f ( x ) = x 2 − 4 x + 5 , x ∈ [ 3 , 4 ]
x v = − − 4 2 ( 1 ) = 2 x_v=-\dfrac{-4}{2(1)}=2 x v = − 2 ( 1 ) − 4 = 2 The function f f f increases on ( 3 , 4 ) (3, 4) ( 3 , 4 )
f ( 3 ) = ( 3 ) 2 − 4 ( 3 ) + 5 = 2 f(3)=(3)^2−4(3)+5=2 f ( 3 ) = ( 3 ) 2 − 4 ( 3 ) + 5 = 2
f ( 4 ) = ( 4 ) 2 − 4 ( 4 ) + 5 = 5 f(4)=(4)^2−4(4)+5=5 f ( 4 ) = ( 4 ) 2 − 4 ( 4 ) + 5 = 5 Domain: [ 3 , 4 ] [3, 4] [ 3 , 4 ]
Range: [ 2 , 5 ] [2, 5] [ 2 , 5 ]
y = x 2 − 4 x + 5 , 3 ≤ x ≤ 4 y=x^2-4x+5, 3\le x\le4 y = x 2 − 4 x + 5 , 3 ≤ x ≤ 4 Change x x x and y y y
x = y 2 − 4 y + 5 , 3 ≤ y ≤ 4 x=y^2-4y+5, 3\le y\le 4 x = y 2 − 4 y + 5 , 3 ≤ y ≤ 4 Solve for y y y
y 2 − 4 y + 4 = x − 1 y^2-4y+4=x-1 y 2 − 4 y + 4 = x − 1 ( y − 2 ) 2 = x − 1 (y-2)^2=x-1 ( y − 2 ) 2 = x − 1 Since 3 ≤ y ≤ 4 3\le y\le 4 3 ≤ y ≤ 4
y − 2 = x − 1 y-2=\sqrt{x-1} y − 2 = x − 1
y = 2 + x − 1 y=2+\sqrt{x-1} y = 2 + x − 1
Then
f − 1 ( x ) = 2 + x − 1 f^{-1}(x)=2+\sqrt{x-1} f − 1 ( x ) = 2 + x − 1
Domain: [ 2 , 5 ] [2, 5] [ 2 , 5 ]
Range: [ 3 , 4 ] [3, 4] [ 3 , 4 ]
3.
f ( x ) = 4 x x − 2 x f(x)=4x\sqrt{x}-2\sqrt{x} f ( x ) = 4 x x − 2 x Domain: [ 0 , ∞ ) [0, \infin) [ 0 , ∞ )
f ′ ( x ) = 4 ( 3 2 ) x − 2 2 x = 6 x − 1 x f'(x)=4(\dfrac{3}{2})\sqrt{x}-\dfrac{2}{2\sqrt{x}}=\dfrac{6x-1}{\sqrt{x}} f ′ ( x ) = 4 ( 2 3 ) x − 2 x 2 = x 6 x − 1
s l o p e = f ′ ( x ) = 6 x − 1 x = 1 slope=f'(x)=\dfrac{6x-1}{\sqrt{x}}=1 s l o p e = f ′ ( x ) = x 6 x − 1 = 1
6 x − x − 1 = 0 6x-\sqrt{x}-1=0 6 x − x − 1 = 0
( 3 x + 1 ) ( 2 x − 1 ) = 0 (3\sqrt{x}+1)(2\sqrt{x}-1)=0 ( 3 x + 1 ) ( 2 x − 1 ) = 0 Since x ≥ 0 , \sqrt{x}\ge0, x ≥ 0 , we take 2 x − 1 = 0 2\sqrt{x}-1=0 2 x − 1 = 0
x = 1 2 \sqrt{x}=\dfrac{1}{2} x = 2 1
x = 1 4 x=\dfrac{1}{4} x = 4 1
f ( 1 4 ) = 4 ( 1 4 ) 1 4 − 2 1 4 = − 1 2 f(\dfrac{1}{4})=4(\dfrac{1}{4})\sqrt{\dfrac{1}{4}}-2\sqrt{\dfrac{1}{4}}=-\dfrac{1}{2} f ( 4 1 ) = 4 ( 4 1 ) 4 1 − 2 4 1 = − 2 1 The tangent line to the graph is
y + 1 2 = x − 1 4 y+\dfrac{1}{2}=x-\dfrac{1}{4} y + 2 1 = x − 4 1
y = x − 3 4 y=x-\dfrac{3}{4} y = x − 4 3
4.
f ( x ) = e x + 3 x 2 + 2 x + 6 f(x)=e^x+3x^2+2x+6 f ( x ) = e x + 3 x 2 + 2 x + 6 Domain: ( − ∞ , ∞ ) (-\infin, \infin) ( − ∞ , ∞ )
f ′ ( x ) = e x + 6 x + 2 f'(x)=e^x+6x+2 f ′ ( x ) = e x + 6 x + 2
f ′ ( x ) = 0 = > e x + 6 x + 2 = 0 f'(x)=0=>e^x+6x+2=0 f ′ ( x ) = 0 => e x + 6 x + 2 = 0
( e x + 6 x + 2 ) ′ = e x + 6 > 0 , x ∈ R (e^x+6x+2)'=e^x+6>0, x\in \R ( e x + 6 x + 2 ) ′ = e x + 6 > 0 , x ∈ R The function g ( x ) = e x + 6 x + 2 g(x)=e^x+6x+2 g ( x ) = e x + 6 x + 2 is strictly increasing on ( − ∞ , ∞ ) . (-\infin, \infin). ( − ∞ , ∞ ) .
The equation e x + 6 x + 2 = 0 e^x+6x+2=0 e x + 6 x + 2 = 0 has the only solution x ≈ − 0.4406. x\approx-0.4406. x ≈ − 0.4406.
The function f ( x ) f(x) f ( x ) is monotonic decreasing on ( − ∞ , − 0.4406 ) . (-\infin, -0.4406). ( − ∞ , − 0.4406 ) .
The function f ( x ) f(x) f ( x ) is monotonic increasing on ( − 0.4406 , ∞ ) . ( -0.4406, \infin). ( − 0.4406 , ∞ ) .
5.
∫ 6 x 3 e x 2 + 2 d x \int 6x^3e^{x^2+2}dx ∫ 6 x 3 e x 2 + 2 d x t = x 2 , d t = 2 x d x t=x^2, dt=2xdx t = x 2 , d t = 2 x d x
∫ 6 x 3 e x 2 + 2 d x = 3 e 2 ∫ t e t d t \int 6x^3e^{x^2+2}dx=3e^2\int te^tdt ∫ 6 x 3 e x 2 + 2 d x = 3 e 2 ∫ t e t d t
∫ t e t d t \int te^t dt ∫ t e t d t
u = t , d u = d t u=t, du=dt u = t , d u = d t
d v = e t d t , v = e t dv=e^tdt, v=e^t d v = e t d t , v = e t
∫ t e t d t = t e t − ∫ e t d t = t e t − e t + C 1 \int te^t dt=te^t -\int e^t dt=te^t-e^t+C_1 ∫ t e t d t = t e t − ∫ e t d t = t e t − e t + C 1
∫ 6 x 3 e x 2 + 2 d x = 3 e x 2 + 2 ( x 2 − 1 ) + C \int 6x^3e^{x^2+2}dx=3e^{x^2+2}(x^2-1)+C ∫ 6 x 3 e x 2 + 2 d x = 3 e x 2 + 2 ( x 2 − 1 ) + C
6.
x 2 + 1 + y ′ cos ( y ) = 0 x^2+1+y'\cos(y)=0 x 2 + 1 + y ′ cos ( y ) = 0
cos ( y ) d y = − ( x 2 + 1 ) d x \cos(y)dy=-(x^2+1)dx cos ( y ) d y = − ( x 2 + 1 ) d x Integrate
∫ cos ( y ) d y = − ∫ ( x 2 + 1 ) d x \int \cos(y)dy=-\int (x^2+1)dx ∫ cos ( y ) d y = − ∫ ( x 2 + 1 ) d x
sin ( y ) = − x 3 3 − x + C \sin (y)=-\dfrac{x^3}{3}-x+C sin ( y ) = − 3 x 3 − x + C
sin ( y ) + x 3 3 + x = C \sin (y)+\dfrac{x^3}{3}+x=C sin ( y ) + 3 x 3 + x = C
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