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sammy [17]
3 years ago
9

Can someone explain to me what a “derivative” means? How do you find the derivative of f(x)=x^3+1?

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
4 0
The derivative is the rate of change of a function, basically represents the slope at different points. To find the derivative of the given function you can use the power rule, which means, if n is a real number, d/dx(x^n)= nx^(n-1). This is a simplification of the chain rule based on the fact that d/dx(x)=1. Anyway, this means that d/dx(x^3 + 1)= 3x^2. Here n is 3 and so it is 3*x^(3-1)= 3x^2. The derivative of x^3+1 is 3x^2.

If you are wondering what happened to the 1, for any constant C, d/dx(C)=0.
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3 years ago
Which equation represents a line that passes through (4,1/3) and has a slope of 3/4?
Alexxandr [17]

Answer:

\large\boxed{B.\ y-\dfrac{1}{3}=\dfrac{3}{4}(x-4)}

Step-by-step explanation:

The point-slope form of an equation:

y-y_1=m(x-x_1)

m - slope

We have

m=\dfrac{3}{4},\ \left(4,\ \dfrac{1}{3}\right)

Substitute:

y-\dfrac{1}{3}=\dfrac{3}{4}(x-4)

7 0
3 years ago
Which of the following pairs of points both lie on the line whose equation is 3x-y=2?
zalisa [80]
Though you did not list the points, I can tell you how to solve for the question.

One way to tell if a point lies on a given line is to take the point and plug it into the equation. If the equation remains true, then the point lies on the line. For example:

If we have the point (1,1), we can plug in 1 for x and 1 for y and see if the equation is true:


3 0
3 years ago
GIVING BRAINLIEST
alexdok [17]
The only reasonable answer is B.
4 0
3 years ago
Read 2 more answers
What is wrong with the following equation?x2+x−20/x−4=x+5a. (x−4)(x−5)≠x2+x−20b. The left-hand side is not defined f
joja [24]

Answer:

The answer to this question can be defined as follows:

In part (i), the answer is "option d".

In part (ii), the answer is "option 2".

Step-by-step explanation:

Given:

Part (a)

\Rightarrow \bold{\frac{x^2+x-20}{x-4}=x+5}\\\\

Solve  the above equation:

\Rightarrow x^2+x-20=(x+5)(x-4)\\\\\Rightarrow x^2+x-20=(x^2-4x+5x-20)\\\\\Rightarrow x^2+x-20=x^2-4x+5x-20\\\\\Rightarrow \boxed{x^2+x-20=x^2+x-20}\\

Given:

Part (b)

\Rightarrow \bold{ \lim_{x \to \3} \frac{x^2+x-12}{x-3}= \lim_{x \to 3} (x+4)}\\\\

Solve the above equation:

factor of \Rightarrow x^2+x-20 =(x-3)(x+4)

\Rightarrow  \lim_{x \to \3} \frac{(x-3)(x+4)}{x-3}= \lim_{x \to 3} (x+4)\\\\\Rightarrow  \lim_{x \to \3} (x+4)= \lim_{x \to 3} (x+4)\\\\

apply limit value:

\Rightarrow  (3+4)=  (3+4)\\\\\Rightarrow  7= 7

6 0
3 years ago
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