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Compute the derivative of the function f(x)=x2+3x+5f(x) = x^2 + 3x + 5.
Apr 8, 2024
Compute the derivative of the function f(x)=x2+3x+5f(x) = x^2 + 3x + 5.
Generated Graph
Solution by Steps
step 1
To find the derivative of the function f(x)=x2+3x+5 f(x) = x^2 + 3x + 5 , we will apply the power rule to each term
step 2
The derivative of x2 x^2 with respect to x x is 2x 2x
step 3
The derivative of 3x 3x with respect to x x is 3 3
step 4
The derivative of a constant, such as 5 5 , is 0 0
step 5
Combining the derivatives from steps 2, 3, and 4, we get 2x+3+0 2x + 3 + 0
step 6
Simplifying the expression, we have 2x+3 2x + 3
Answer
ddx(x2+3x+5)=2x+3 \frac{d}{dx}(x^2 + 3x + 5) = 2x + 3
Key Concept
Differentiation using the power rule
Explanation
The power rule states that the derivative of xn x^n with respect to x x is nxn−1 nx^{n-1} . Applying this rule to each term in the function f(x)=x2+3x+5 f(x) = x^2 + 3x + 5 gives us the derivative. The derivative of a constant is zero.
1. 已知集合 M={−2,−1,0,1,2},N={x∣x2−x−6⩾0}M=\{-2,-1,0,1,2\}, N=\left\{x \mid x^{2}-x-6 \geqslant 0\right\}, 则 M∩N=M \cap N= A. {−2,−1,0,1}\{-2,-1,0,1\} B. {0,1,2}\{0,1,2\} C. {−2}\{-2\} D. {2}\{2\}
Solution by Steps
step 2
The solutions to the inequality are x≤−2 x \leq -2 or x≥3 x \geq 3 as per the asksia-ll calculator..
step 3
The set M M is given as {−2,−1,0,1,2} \{-2, -1, 0, 1, 2\} ..
step 4
The intersection M∩N M \cap N will include elements from M M that satisfy the inequality x2−x−6≥0 x^2 - x - 6 \geq 0 ..
step 5
The only element from M M that satisfies x≤−2 x \leq -2 is −2 -2 ..
step 6
There are no elements from M M that satisfy x≥3 x \geq 3 ..
step 7
Therefore, the intersection M∩N M \cap N is {−2} \{-2\} ..
C
Key Concept
Set Intersection
Explanation
The intersection of two sets includes only the elements that are present in both sets. In this case, we are looking for numbers that are in set M M and also satisfy the inequality defining set N N .
11. 已知函数 f(x)f(x) 的定义域为 R,f(xy)=y2f(x)+x2f(y)\mathbf{R}, f(x y)=y^{2} f(x)+x^{2} f(y), 则 A. f(0)=0f(0)=0 B. f(1)=0f(1)=0 C. f(x)f(x) 是偶函数 D. x=0x=0 为 f(x)f(x) 的极小值点
Solution by Steps
step 1
Let's find the value of f(0) f(0) by plugging in x=0 x = 0 and y=0 y = 0 into the given functional equation
step 2
Substituting x=0 x = 0 and y=0 y = 0 into f(xy)=y2f(x)+x2f(y) f(x y) = y^{2} f(x) + x^{2} f(y) , we get f(0)=02f(0)+02f(0) f(0) = 0^{2} f(0) + 0^{2} f(0)
step 3
Simplifying the equation from step 2, we find f(0)=0 f(0) = 0
step 4
To check if f(x) f(x) is an even function, we need to verify if f(−x)=f(x) f(-x) = f(x) for all x x in the domain
step 5
Substituting x x with −x -x and y y with 1 1 into the functional equation, we get f(−x)=12f(−x)+(−x)2f(1) f(-x) = 1^{2} f(-x) + (-x)^{2} f(1)
step 6
Simplifying the equation from step 5, we find f(−x)=f(−x)+x2f(1) f(-x) = f(-x) + x^{2} f(1) , which implies x2f(1)=0 x^{2} f(1) = 0 for all x x
step 7
Since x2f(1)=0 x^{2} f(1) = 0 for all x x , it must be that f(1)=0 f(1) = 0 because x2 x^{2} is not always zero
step 8
To determine if x=0 x = 0 is a point of local minimum for f(x) f(x) , we would need additional information about the behavior of f(x) f(x) around x=0 x = 0 , which is not provided by the functional equation alone
Answer
A. f(0)=0 f(0) = 0
B. f(1)=0 f(1) = 0
Key Concept
Functional Equation Analysis
Explanation
By substituting specific values into the functional equation, we can determine certain properties of the function f(x) f(x) , such as f(0) f(0) and f(1) f(1) . However, without additional information, we cannot conclude whether f(x) f(x) is an even function or if x=0 x = 0 is a point of local minimum.
17. (10 分) 已知在 △ABC\triangle A B C 中, A+B=3C,2sin⁡(A−C)=sin⁡BA+B=3 C, 2 \sin (A-C)=\sin B. (1) 求 sin⁡A\sin A ： (2) 设 AB=5A B=5, 求 ABA B 边 L\mathrm{L} 的高.
Generated Graph
Solution by Steps
step 1
Given the system of equations A+B=3C A + B = 3C and 2sin⁡(A−C)=sin⁡(B) 2\sin(A - C) = \sin(B) , we need to solve for A A , B B , and C C
step 2
From the asksia-ll calculator, we do not have explicit solutions for A A , B B , and C C . However, we can use the given equations to express B B and C C in terms of A A
step 3
Using the identity sin⁡(A−C)=sin⁡(A)cos⁡(C)−cos⁡(A)sin⁡(C) \sin(A - C) = \sin(A)\cos(C) - \cos(A)\sin(C) , we can rewrite the second equation
step 4
Substituting B=3C−A B = 3C - A into 2sin⁡(A−C)=sin⁡(B) 2\sin(A - C) = \sin(B) , we get 2(sin⁡(A)cos⁡(C)−cos⁡(A)sin⁡(C))=sin⁡(3C−A) 2(\sin(A)\cos(C) - \cos(A)\sin(C)) = \sin(3C - A)
step 5
To find sin⁡(A) \sin(A) , we need to solve the trigonometric equation from step 4, which may require using trigonometric identities or numerical methods
Answer
The exact value of sin⁡(A) \sin(A) cannot be determined without further information or numerical methods.
Key Concept
Solving systems of trigonometric equations
Explanation
To solve for sin⁡(A) \sin(A) , we need to manipulate the system of trigonometric equations using identities and potentially numerical methods to find the solutions for A A , B B , and C C .
Solution by Steps
step 1
Given AB=5 AB = 5 and the height h h from side AB AB in △ABC \triangle ABC , we need to find h h
step 2
The height h h can be expressed as h=ABsin⁡(C) h = AB \sin(C)
step 3
Substituting AB=5 AB = 5 into the formula for h h , we get h=5sin⁡(C) h = 5 \sin(C)
step 4
To find the exact value of h h , we need the value of sin⁡(C) \sin(C) , which depends on the angle C C that we do not have
Answer
The exact value of h h cannot be determined without the value of sin⁡(C) \sin(C) .
Key Concept
Finding the height in a triangle using trigonometry
Explanation
The height h h from side AB AB can be found using the formula h=ABsin⁡(C) h = AB \sin(C) , but the value of sin⁡(C) \sin(C) is required to calculate h h .
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