What is this asking I have no idea
First, you need to find the derivative of this function. This is done by multiplying the exponent of the variable by the coefficient, and then reducing the exponent by 1.
f'(x)=3x^2-3
Now, set this function equal to 0 to find x-values of the relative max and min.
0=3x^2-3
0=3(x^2-1)
0=3(x+1)(x-1)
x=-1, 1
To determine which is the max and which is the min, plug in values to f'(x) that are greater than and less than each. We will use -2, 0, 2.
f'(-2)=3(-2)^2-3=3(4)-3=12-3=9
f'(0)=3(0)^2-3=3(0)-3=0-3=-3
f'(2)=3(2)^2=3(4)-3=12-3=9
We examine the sign changes to determine whether it is a max or a min. If the sign goes from + to -, then it is a maximum. If it goes from - to +, it is a minimum. Therefore, x=-1 is a relative maximum and x=1 is a relative miminum.
To determine the values of the relative max and min, plug in the x-values to f(x).
f(-1)=(-1)^3-3(-1)+1=-1+3+1=3
f(1)=(1)^3-3(1)+1=1-3+1=-1
Hope this helps!!
Answer:
Step-by-step explanation:
The slope-intercept form of a straight line equation is y = mx + c, where m is the slope and c is the y-intercept of the line.
Now, we know that if two straight lines are perpendicular to each other then the product of their slopes will be -1.
So, the equation of a straight line which is perpendicular to the line 
will be 
....... (1), where c' is constant.
Given that the line (1) passes through (2,4) point.
Hence, 
⇒ c' = 1.
Therefore, the final equation of the required straight line is . (Answer)
A. y=-4x-4 where m=-4 and c=-4 for the equation y=mx+c
