As you progress in math, it will become increasingly important that you know how to express exponentiation properly.
y = 2x3 – x2 – 4x + 5 should be written <span>y = 2^x3 – x^2 – 4^x + 5. The
" ^ " symbol denotes exponentiation.
I see you're apparently in middle school. Is that so? If so, are you taking calculus already? If so, nice!
Case 1: You do not yet know calculus and have not differentiated or found critical values. Sketch the function </span>y = 2x^3 – x^2 – 4^x + 5, including the y-intercept at (0,5). Can you identify the intervals on which the graph appears to be increasing and those on which it appears to be decreasing?
Case 2: You do know differentiation, critical values and the first derivative test. Differentiate y = 2x^3 – x^2 – 4^x + 5 and set the derivative = to 0:
dy/dx = 6x^2 - 2x - 4 = 0. Reduce this by dividing all terms by 2:
dy/dx = 3x^2 - x - 2 = 0 I used synthetic div. to determine that one root is x = 2/3. Try it yourself. This leaves the coefficients of the other factor, (3x+3); this other factor is x = 3/(-3) = -1. Again, you should check this.
Now we have 2 roots: -1 and 2/3
Draw a number line. Locate the origin (0,0). Plot the points (-1, 0) and (2/3, 0). This subdivides the number line into 3 subintervals:
(-infinity, -1), (-1, 2/3) and (2/3, infinity).
Choose a test number from each interval and subst. it for x in the derivative formula above. If the derivative comes out +, the function is increasing on that interval; if -, the function is decreasing.
Ask all the questions you want, if this explanation is not sufficiently clear.
Answer:
It is 2/7 please mark me brainliest
Step-by-step explanation:
Number 2 for question 1 and number 4 for question 2
Answer:
The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Step-by-step explanation:
Given : The position of an object along a vertical line is given by 
, where s is measured in feet and t is measured in seconds.
To find : What is the maximum velocity of the object in the time interval [0, 4]?
Solution :
The velocity is rate of change of distance w.r.t time.
Distance in terms of t is given by,
Derivate w.r.t. time,
It is a quadratic function so its maximum is at vertex of the function.
The x point of the function is given by,
Where, a=-3, b=6 and c=7
As 1 lie between interval [0,4]
Substitute t=1 in the function,
Th maximum velocity is 10 ft/s.
Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
