cho hàm số y = f(x) = 2x - 3
Tính X biết y = \(\dfrac{-1}{3}\)
Câu 1: Cho hàm số y = 2x\(^2\)
a) Hãy lập bảng tính các giá trị f(-5), f(-3), f(0), f(3), f(5)
b) Tìm x biết f(x) = 8, f(x) = 6 - 4\(\sqrt{2}\)
Câu 2: Cho hàm số y = f(x) = \(\dfrac{1}{3}x^2\)
Tìm các giá trị của x, biết rằng \(y=\dfrac{1}{27}\). Cũng câu hỏi tương tự với y = 5
Câu 1:
a)
\(y=f\left(x\right)=2x^2\) | -5 | -3 | 0 | 3 | 5 |
f(x) | 50 | 18 | 0 | 18 | 50 |
b) Ta có: f(x)=8
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
Vậy: Để f(x)=8 thì \(x\in\left\{2;-2\right\}\)
Ta có: \(f\left(x\right)=6-4\sqrt{2}\)
\(\Leftrightarrow2x^2=6-4\sqrt{2}\)
\(\Leftrightarrow x^2=3-2\sqrt{2}\)
\(\Leftrightarrow x=\sqrt{3-2\sqrt{2}}\)
hay \(x=\sqrt{2}-1\)
Vậy: Để \(f\left(x\right)=6-4\sqrt{2}\) thì \(x=\sqrt{2}-1\)
: Cho hàm số y = f(x) = -2x + 3.
a) Tính f(-2); f(0); f(-\(\dfrac{1}{2}\)). b) Tìm các giá trị của x biết : f(x) = 5 ; f(x) = 1
a) Cho hàm số y = f(x) = -2x + 3.
Ta có: f(-2)= -2.(-2)+3
= 4+3=7
Ta có: f(0)= -2.0+3
= 0+3=3
Ta có: f(
Lời giải:
a.
$f(-2)=(-2)(-2)+3=7$
$f(0)=(-2).0+3=3$
$f(\frac{-1}{2})=(-2).\frac{-1}{2}+3=4$
b.
$f(x)=-2x+3=5$
$\Rightarrow -2x=2$
$\Rightarrow x=-1$
$f(x)=-2x+3=1$
$\Rightarrow -2x=1-3=-2$
$\Rightarrow x=1$
Bài 1: Xét tính đơn điệu của hàm số \(y=f(x)\) khi biết đạo hàm của hàm số là:
a) \(f'(x)=(x+1)(1-x^2)(2x-1)^3\)
b) \(f'(x)=(x+2)(x-3)^2(x-4)^3\)
Bài 2: Cho hàm số \(y=f(x)\) có đạo hàm \(f'(x)=x(x+1)(x-2)\). Xét tính biến thiên của hàm số:
a) \(y=f(2-3x)\)
b) \(y=f(x^2+1)\)
c) \(y=f(3x+1)\)
tìm x:
(1)
a) \(x+\dfrac{2}{3}=\dfrac{-1}{12}\)
b)\(\left(2x+1\right)^2=9\)
(2) cho hàm số y=f(x)=2x2+4. Tính f(2);f(-1)
(1)
a) x=\(\dfrac{-1}{12}-\dfrac{2}{3}\)=\(\dfrac{-3}{4}\)
b) 2x+1=3 => 2x=3-1=2 => x=1
(2)
f(2)=2.22+4=12
f(-1)=2.(-1)2+4=6
(1)
a) \(x+\dfrac{2}{3}=-\dfrac{1}{12}\\ \Rightarrow x=-\dfrac{1}{12}-\dfrac{2}{3}\\ \Rightarrow x=\dfrac{-1}{12}-\dfrac{8}{12}\\ \Rightarrow x=-\dfrac{9}{12}=-\dfrac{3}{4}\)
Vậy \(x=-\dfrac{3}{4}\)
b) \(\left(2x+1\right)^2=9\\ \Rightarrow\left(2x+1\right)^2=3^2=\left(-3\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=2\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-2;1\right\}\)
(2)
\(y=f\left(x\right)=2x^2+4\\ f\left(2\right)=2\cdot2^2+4=8+4=12\\ f\left(-1\right)=2\cdot\left(-1\right)^2+4=2+4=6\)
Vậy \(f\left(2\right)=12\\ f\left(-1\right)=6\)
Tính đạo hàm của các hàm số sau:
a) \(y = (2x^2 - x + 1)^{\frac{1}{3}}\)
b) \(y = (3x+1)^{\pi}\)
c) \(y = \sqrt[3]{\dfrac{1}{x-1}}\)
d) \(y =\log_{3} \left(\dfrac{x+1}{x-1}\right)\)
e) \(y = 3^{x^{2}}\)
f) \(y = \left(\dfrac{1}{2}\right)^{x^2-1}\)
h) \(y = (x+1) . e^{cosx}\)
g) \(y = \ln (x^2+x+1)\)
l) \(y = \dfrac{\ln x}{x+1}\)
a: \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
=>\(y'=\dfrac{1}{3}\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}\cdot\left(2x^2-x+1\right)'\)
\(=\dfrac{1}{3}\cdot\left(4x-1\right)\left(2x^2-x+1\right)^{-\dfrac{2}{3}}\)
b: \(y=\left(3x+1\right)^{\Omega}\)
=>\(y'=\Omega\cdot\left(3x+1\right)'\cdot\left(3x+1\right)^{\Omega-1}\)
=>\(y'=3\Omega\left(3x+1\right)^{\Omega-1}\)
c: \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
=>\(y'=\dfrac{\left(\dfrac{1}{x-1}\right)'}{3\cdot\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(=\dfrac{\dfrac{1'\left(x-1\right)-\left(x-1\right)'\cdot1}{\left(x-1\right)^2}}{\dfrac{3}{\sqrt[3]{\left(x-1\right)^2}}}\)
\(=\dfrac{-x}{\left(x-1\right)^2}\cdot\dfrac{\sqrt[3]{\left(x-1\right)^2}}{3}\)
\(=\dfrac{-x}{\sqrt[3]{\left(x-1\right)^4}\cdot3}\)
d: \(y=log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Leftrightarrow y'=\dfrac{\left(\dfrac{x+1}{x-1}\right)'}{\dfrac{x+1}{x-1}\cdot ln3}\)
\(\Leftrightarrow y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}:\dfrac{ln3\left(x+1\right)}{x-1}\)
\(\Leftrightarrow y'=\dfrac{x-1-x-1}{\left(x-1\right)^2}\cdot\dfrac{x-1}{ln3\cdot\left(x+1\right)}\)
\(\Leftrightarrow y'=\dfrac{-2}{\left(x-1\right)\cdot\left(x+1\right)\cdot ln3}\)
e: \(y=3^{x^2}\)
=>\(y'=\left(x^2\right)'\cdot ln3\cdot3^{x^2}=2x\cdot ln3\cdot3^{x^2}\)
f: \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
=>\(y'=\left(x^2-1\right)'\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}=2x\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}\)
h: \(y=\left(x+1\right)\cdot e^{cosx}\)
=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)
=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)
\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)
a) \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}.\left(4x-1\right)\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{-\dfrac{2}{3}}.\left(4x-1\right)\)
b) \(y=\left(3x+1\right)^{\pi}\)
\(\Rightarrow y'=\pi.\left(3x+1\right)^{\pi-1}.3=3\pi.\left(3x+1\right)^{\pi-1}\)
c) \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
\(\Rightarrow y'=\dfrac{\left(x-1\right)^{-1-1}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^{3-1}}}=\dfrac{\left(x-1\right)^{-2}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}=\dfrac{1}{3.\sqrt[]{x-1}.\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(\Rightarrow y'=\dfrac{1}{3\left(x-1\right)^{\dfrac{1}{2}}.\left(x-1\right)^{\dfrac{2}{3}}}=\dfrac{1}{3\left(x-1\right)^{\dfrac{7}{6}}}=\dfrac{1}{3\sqrt[6]{\left(x-1\right)^7}}\)
d) \(y=\log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Rightarrow y'=\dfrac{\dfrac{1-\left(-1\right)}{\left(x-1\right)^2}}{\dfrac{x+1}{x-1}.\ln3}=\dfrac{2}{\left(x+1\right)\left(x-1\right).\ln3}\)
e) \(y=3^{x^2}\)
\(\Rightarrow y'=3^{x^2}.ln3.2x=2x.3^{x^2}.ln3\)
f) \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
\(\Rightarrow y'=\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}.2x=2x.\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}\)
Các bài còn lại bạn tự làm nhé!
2
a.cho hàm số y=f(x)=\(\dfrac{2}{3}x\).Tính f(-2),f(-1),f(0),f(\(\dfrac{1}{2}\)),f(1),f(2),f(3).
b,
cho hàm số y=g(x)=\(\dfrac{2}{3}x\)+3.Tính g(-2),g(-1),g(0),g(\(\dfrac{1}{2}\)),g(1),g(2),g(3)
c.có nhận xét gì về giá trị 2 hàm số đã cho ở trên khi biến x lấy cùng 1 giá trị
c: Ở hai hàm số trên, nếu lấy biến x cùng một giá trị thì f(x) sẽ nhỏ hơn g(x) 3 đơn vị
1. Tính đạo hàm của các hàm số sau:
a, \(y=\dfrac{2x-1}{x-1}\)
b, \(y=\dfrac{2x+1}{1-3x}\)
c, \(y=\dfrac{x^2+2x+2}{x+1}\)
d, \(y=\dfrac{2x^2}{x^2-2x-3}\)
e, \(y=x+1-\dfrac{2}{x-1}\)
g, \(y=\dfrac{2x^2-4x+5}{2x+1}\)
2. Tính đạo hàm của các hàm số sau:
a, \(y=\left(x^2+x+1\right)^4\)
b, y= (1-2x2)5
c, \(y=\left(\dfrac{2x+1}{x-1}\right)^3\)
d, \(y=\dfrac{\left(x+1\right)^2}{\left(x-1\right)^3}\)
e, \(y=\dfrac{1}{\left(x^2-2x+5\right)^2}\)
f, \(y=\left(3-2x^2\right)^4\)
a. \(y'=\dfrac{-1}{\left(x-1\right)}\)
b. \(y'=\dfrac{5}{\left(1-3x\right)^2}\)
c. \(y=\dfrac{\left(x+1\right)^2+1}{x+1}=x+1+\dfrac{1}{x+1}\Rightarrow y'=1-\dfrac{1}{\left(x+1\right)^2}=\dfrac{x^2+2x}{\left(x+1\right)^2}\)
d. \(y'=\dfrac{4x\left(x^2-2x-3\right)-2x^2\left(2x-2\right)}{\left(x^2-2x-3\right)^2}=\dfrac{-4x^2-12x}{\left(x^2-2x-3\right)^2}\)
e. \(y'=1+\dfrac{2}{\left(x-1\right)^2}=\dfrac{x^2-2x+3}{\left(x-1\right)^2}\)
g. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)
2.
a. \(y'=4\left(x^2+x+1\right)^3.\left(x^2+x+1\right)'=4\left(x^2+x+1\right)^3\left(2x+1\right)\)
b. \(y'=5\left(1-2x^2\right)^4.\left(1-2x^2\right)'=-20x\left(1-2x^2\right)^4\)
c. \(y'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{2x+1}{x-1}\right)'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{-3}{\left(x-1\right)^2}\right)=\dfrac{-9\left(2x+1\right)^2}{\left(x-1\right)^4}\)
d. \(y'=\dfrac{2\left(x+1\right)\left(x-1\right)^3-3\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)^6}=\dfrac{-x^2-6x-5}{\left(x-1\right)^4}\)
e. \(y'=-\dfrac{\left[\left(x^2-2x+5\right)^2\right]'}{\left(x^2-2x+5\right)^4}=-\dfrac{2\left(x^2-2x+5\right)\left(2x-2\right)}{\left(x^2-2x+5\right)^4}=-\dfrac{4\left(x-1\right)}{\left(x^2-2x+5\right)^3}\)
f. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
Cho hàm số y=f(x)=I3x-1I
a) Tính f(-2);f(2);f(\(\dfrac{-1}{4}\))
b)Tìm x biết f(x)=10;f(x)=-3
a) Ta có:
\(f\left(-2\right)=\left|3\cdot-2-1\right|=\left|-6-1\right|=\left|-7\right|=7\)
\(f\left(2\right)=\left|3\cdot2-1\right|=\left|6-1\right|=5\)
\(f\left(-\dfrac{1}{4}\right)=\left|3\cdot-\dfrac{1}{4}-1\right|=\left|-\dfrac{3}{4}-1\right|=\left|-\dfrac{7}{4}\right|=\dfrac{7}{4}\)
b) Ta có:
\(f\left(x\right)=10\)
\(\Rightarrow\left|3x-1\right|=10\)
Với \(x\ge\dfrac{1}{3}\Rightarrow3x-1=10\)
\(\Rightarrow3x=11\Rightarrow x=\dfrac{11}{3}\left(tm\right)\)
Với \(x< \dfrac{1}{3}\Rightarrow3x-1=-10\)
\(\Rightarrow3x=-9\Rightarrow x=-3\left(tm\right)\)
_______
\(f\left(x\right)=-3\)
\(\Rightarrow\left|3x-1\right|=-3\)
Mà: \(\left|3x-1\right|\ge0\forall x\) và \(-3< 0\)
\(\Rightarrow\left|3x-1\right|=-3\) (vô lý)
\(\Rightarrow\) không có x thỏa mãn
tính đạo hàm của các hàm số sau
a) \(y=\dfrac{x^2+3x-1}{x+2}\)
b) \(y=\dfrac{2x^2-x}{x^2+1}\)
c) \(y=\dfrac{3-2x}{x-1}+\sqrt{2x-3}\)
a: \(y'=\dfrac{\left(x^2+3x-1\right)'\cdot\left(x+2\right)-\left(x^2+3x-1\right)\cdot\left(x+2\right)'}{\left(x+2\right)^2}\)
\(=\dfrac{\left(2x+3\right)\left(x+2\right)-\left(x^2+3x-1\right)}{\left(x+2\right)^2}\)
\(=\dfrac{2x^2+7x+6-x^2-3x+1}{\left(x+2\right)^2}=\dfrac{x^2+4x+7}{\left(x+2\right)^2}\)
b: \(y'=\dfrac{\left(2x^2-x\right)'\cdot\left(x^2+1\right)-\left(2x^2-x\right)\left(x^2+1\right)'}{\left(x^2+1\right)^2}\)
\(=\dfrac{4x\left(x^2+1\right)-2x\left(2x^2-x\right)}{\left(x^2+1\right)^2}\)
\(=\dfrac{4x^3+4x-4x^3+2x^2}{\left(x^2+1\right)^2}=\dfrac{2x^2+4x}{\left(x^2+1\right)^2}\)
c: \(\left(\dfrac{3-2x}{x-1}\right)'=\dfrac{\left(3-2x\right)'\left(x-1\right)-\left(3-2x\right)\left(x-1\right)'}{\left(x-1\right)^2}\)
\(=\dfrac{-2\left(x-1\right)-\left(3-2x\right)}{\left(x-1\right)^2}=\dfrac{-2x+2-3+2x}{\left(x-1\right)^2}=-\dfrac{1}{\left(x-1\right)^2}\)
\(\left(\sqrt{2x-3}\right)'=\dfrac{\left(2x-3\right)'}{2\sqrt{2x-3}}=\dfrac{1}{\sqrt{2x-3}}\)
\(y'=\left(\dfrac{3-2x}{x-1}\right)'+\left(\sqrt{2x-3}\right)'\)
\(=\dfrac{-1}{\left(x-1\right)^2}+\dfrac{1}{\sqrt{2x-3}}\)
1. đạo hàm của hàm số f(x) = 2x - 5 tại \(x_0=4\)
2. đạo hàm của hàm số \(y=x^2-3\sqrt{x}+\dfrac{1}{x}\)
3. đạo hàm của hàm số \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt{x}\) tại điểm x = 1
1) \(f\left(x\right)=2x-5\)
\(f'\left(x\right)=2\)
\(\Rightarrow f'\left(4\right)=2\)
2) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)
\(\Rightarrow y'=2x-\dfrac{3}{2\sqrt[]{x}}-\dfrac{1}{x^2}\)
3) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)
\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1.\left(x+9\right)}{\left(x-3\right)^2}+\dfrac{4}{2\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{12}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=2\left[\dfrac{6}{\left(x-3\right)^2}+\dfrac{1}{\sqrt[]{x}}\right]\)
\(\Rightarrow f'\left(1\right)=2\left[\dfrac{6}{\left(1-3\right)^2}+\dfrac{1}{\sqrt[]{1}}\right]=2\left(\dfrac{3}{2}+1\right)=2.\dfrac{5}{2}=5\)