PLEASE ANSWER ALL (7--10) Determine the intervals on which the function is conca

Question

PLEASE ANSWER ALL

(7--10) Determine the intervals on which the function is concave up and intervals on which the function is concave down. . No need to include these graphs.

12) Using the secant definition of concavity, explain that if the function

PLEASE ANSWER ALL (7--10) Determine the intervals on which the function is concave up and intervals on which the function is concave down. . No need to include these graphs. 7) f'(x) > [f(x)-f(a)]/[x-a] . I. Show that f'' > 0 on f is concave up. 13) Please Explain. Assume that a, then x > a is increasing for all s(x) = [f(x)-f(a)]/[x-a] defined for (1,f(1)). 12) Using the secant definition of concavity, explain that if the function (-1,f(-1)), and f'(c) is the slope of the secant line passing through x = c such that x in [-1,1], satisfy the hypotheses of the Mean Value Theorem? Find f(x) = 3 x^2+5 x-2, defined for f(x) = [x^2]/[x^2-1].

Explanation / Answer

7)

y = x^3

y' = 3x^2

y'' = 6x = 0 ---> x = 0

So, this splits the number line into (-inf , 0) and ( 0 ,inf)

Region 1 : (-inf , 0)
Testvalue = -1
y'' = 6x
y'' = 6(-1) = -6 ---> negative
So, concave down

Region 2 : (0 , inf)
Testvalue = 1
y'' = 6x
y'' = 6(1) --> 6 --> positive
So, concave up

Concave up over (0 , inf) --> ANSWER
Concave down over (-inf , 0) ----> ANSWER

---------------------------------------------------------------------------------------

8)

y = xe^(-x)

y' = e^(-x) - xe^(-x)

y'' = -e^(-x) - e^(-x) + xe^(-x)

y'' = xe^(-x) - 2e^(-x) = 0

xe^(-x) - 2e^-x = 0

e^-x*(x - 2) = 0

x = 2 --> This is the only inflection point

This splits number line into (-inf , 2) and (2 , inf)

Region 1 : (-inf , 2)
Testvalue = 0
y'' = xe^(-x) - 2e^(-x)
y'' = 0 - 2
y'' = -2 --> negative
So, concave down

Region 2 : (2 , inf)
Testvalue = 3
y'' = 3e^-3 - 2e^-3 --> positive
So, concave up

Concave up over (2 , inf) --> ANSWER
Concave down over (-inf , 2) ---> ANSWER

--------------------------------------------------------------------------------

9)

y = x - 2sinx

y' = 1 - 2cosx

y'' = 2sinx = 0 ---> sinx = 0 ---> x = pi , 2pi

Region 1 : (0 , pi)
Testvalue = pi/2
y'' = 2sin(x)
2sin(pi/2)
2 --> positive
So, concave up

Region 2 : (pi , 2pi)
Testvalue = 3pi/2
y'' = 2sin(3pi/2)
y'' = -2 --> negative
So, concave down

Region 3 : (2pi , 3pi)
Testvalue = 5pi/2
y'' = 2sin(5pi/2)
y'' = 2 --> positive
So, concave up

So, concave up over (0 , pi) and (2pi , 3pi) -- ANSWER
Concave down over (pi , 2pi) --> ANSWER

----------------------------------------------------------------------------------------

10)

y = x^2 / (x^2 - 1)

y' = -2x / (x^2 - 1)^2

y'' = (6x^2 + 2) / (x^2 - 1)^3 --> no inflection points when we solve this.....

So, only things to check are the vertical asymptotes, which here are x = -1 and x = 1

Region 1 : (-inf , -1)
Testvalue = -2
y'' = (6x^2 + 2) / (x^2 - 1)^3
y'' at x = -2 is : 26/27 --> positive
So, concave up

Region 2 : (-1 , 1)
Testvalue = 0
y'' = -2 --> negative
So, concave down

Region 3 : (1 , inf)
Testvalue = 2
y'' = 26/27 --> positive

So, concave up over (-inf , -1) U (1 , inf) --> ANSWER
Concave down over (-1 , 1) --> ANSWER

------------------------------------------------------------------------------------------------

11)

f(x) = 3x^2 + 5x - 2

The fuction is continuous over [-1 , 1] and differentiable over (-1 , 1)
So, yes, the conditions are satisfied...

f'(x) = 6x + 5

So, f'(c) = 6c + 5

f(1) = 3 + 5 - 2 --> 6
f(-1) = 3 - 5 - 2 --> -4

(f(1) - f(-1)) / (1 - (-1))

(6 - (-4)) / 2

10/2

5

So, equating :

6c + 5 = 5

6c = 0

c = 0

So, the value of x is : x = c = 0

And this value does lie within [-1 , 1], the given interval...

So, c = 0 -----> ANSWER

---------------------------------------------------------------------------

s(x) = (f(x) - f(a)) / (x - a)

This is nothing but the difference quotient over (a , f(a)) and (x , f(x)), which is nothing but the slope of the secant line....

Now, if s(x), the first derivative is an increasing function, then it means that the second derivative is ALWAYS POSITIVE, which in layman's terms indicates "concave up"

Hence proved!

-----------------------------------------------------------------------------------------

f'' > 0 --> so function is concave up everywhere

f'(x) > (f(x) - f(a)) / (x - a)

For the function to be concave up, the derivative, which is the slope of tangent line must ALWAYS be greater than the slope of the secant line at that interval

Now, secant line has slope = (f(x) - f(a)) / (x - a)

Tangent line slope = first derivative = f'(x)

So, if tangent line slope > secant line slope then, we get :

f'(x) > (f(x) - f(a)) / (x - a)

Hence proved!

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.