Plz help x -3 0 3 4

f(x) 5 -1 2 7

f'(x) -2 4 0 1

The table above gives values of a twice-differentiable function f and its first derivative f′ for selected values of x . Let g be the function defined by g(x)=f(2x−x2) .

(a) What is the value of g′(−1) ?

(b) It is known that g′′(0)=0. What is the value of f′′(0) ?

(c) Is there a value c, for 0
(d) Let h be the function with first derivative given by h′(x)=4xex. At what value of x in the interval 0≤x≤4 does the instantaneous rate of change of h equal the average rate of change of f over the interval 0≤x≤4 ?

$79500

step-by-step explanation:

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)

b and c are the correct answers.

good luck on the quiz! : d

here is me explaining it:

the pizza would be delivered around 5, give or take 15 minutes. this means it could be delivered anywhere from 4: 45 to 5: 15

let's say 5: 00 is equivalent to 0 on a number line and x represents time. 4: 45 would be equal to -15x because we are 15 minutes behind our starting point, 5: 00. 5: 15 would be equal to 15x because we are 15 minutes ahead of your starting point (which means you're late on the pizza delivery! )

absolute value is very confusing. the basic definition is how far you are from your starting point. if we deliver the pizza at 4: 45 (15 minutes before the starting point) we would be -15. this means we are 15 away from our starting point, 5: 00. that's the absolute value. |-15| (absolute value of negative fifteen) is equal to 15. if you did not understand this i suggest rereading this paragraph.

the normal way you would write this scenario would be 4.75 (4 and 3 quarters, or 4: 45. you can also write as a fraction.) is less than x and x is less than 5.25 (5 and 1 quarter or 5: 15). 4.75< x< 5.25 the reason why i am using numbers is because you cannot represent typical time as values on a graph, number line, or in an equation.

another way to write this scenario is with absolute value. we can write this as |x| being less than or equal to .25 (remember, this just means 15 minutes).           |x| ≤ .25.

sorry if the end was confusing. because we are talking about time this whole question is 10x harder. like i said earlier, you cannot represent time as, well, time, in a graph, equation, or number line. i set the values as follows: 1/4 or .25 for 15 minutes, and 1 for 1 hour. this is just a common way to set up graphs involving time.

|x| ≤ .25

i hope this ! : )

tell me if you need more assistance