The table below gives selected values of a twice differentiable function f(x)

x|. -7. -6. -4. -2.

f(x)|. 0. -1. -2. 0

f'(x)|. 3. 2. -1. 7

What is the limit of f(3x-1)/(x^2)-4 as x approaches -2?

What is the limit of f(f(x))/5x + 20 as x approaches -4?

f(3(-2)-1)/(-2)^2 - 4 = f(-7)/4 - 4 = 0 - 4 = -4

Now, if you meant
f(3x-1)/(x^2-4) then that would be 0/0 which is a problem.
But, l'Hôpital's Rule says that the limit is the same as
d/dx f(3x-1) / d/dx (x^2-4)
= (f'(3x-1) * 3) / (2x)
= (3*3) / -6
= -3/2

see what you can do with the other one

To find the limits of the given expressions, we can use the properties of limits and the given information about the function and its derivative.

1. The limit of f(3x-1)/(x^2)-4 as x approaches -2:

Let's substitute the given values of x into the expression and see if we can observe a pattern:

x: -7 -6 -4 -2
f(x): 0 -1 -2 0

From the values we see that when x approaches -2, f(x) also approaches 0. Therefore, we can substitute f(x) = 0 into the expression:

lim(x->-2) [f(3x-1)/(x^2)-4] = [0/(0^2)-4] = -4

So, the limit of f(3x-1)/(x^2)-4 as x approaches -2 is -4.

2. The limit of f(f(x))/5x + 20 as x approaches -4:

Again, let's observe the values of f(x) for the given x-values:

x: -7 -6 -4 -2
f(x): 0 -1 -2 0

We can notice that when x approaches -4, f(x) approaches -2. Therefore, we can substitute f(x) = -2 into the expression:

lim(x->-4) [f(f(x))/5x + 20] = [f(-2)/(5(-4))+20] = [-2/(5(-4))+20]

Now, let's find f(-2) by using the given derivative values:

f'(-2) = 7

Since f'(x) gives the rate of change of f(x), we can find f(x) by integrating f'(x):

∫ f'(x) dx = ∫ 7 dx
f(x) = 7x + C

Using the point (-2, 0), we can solve for C:

0 = 7(-2) + C
C = 14

So, f(x) = 7x + 14. Now, substitute f(x) = -2 into the expression:

lim(x->-4) [f(f(x))/5x + 20] = [f(-2)/(5(-4))+20] = [-2/(5(-4))+20]
= [-2/(-20)+20]
= [1/10 + 20]
= [1/10 + 200/10]
= [201/10]

Therefore, the limit of f(f(x))/5x + 20 as x approaches -4 is 201/10.

To find the limit of a function as x approaches a certain value, we can use the concept of continuity. If a function is continuous at a given point, then the limit at that point can simply be evaluated by plugging in the value. However, if the function is not continuous at that point, we need to employ other methods.

For the first question, the limit of f(3x-1)/(x^2)-4 as x approaches -2, we can directly substitute -2 into the expression:

f(3(-2)-1)/((-2)^2)-4

= f(-7)/4-4

Using the given values in the table:

= 0/4-4

= 0-4

= -4

Therefore, the limit of f(3x-1)/(x^2)-4 as x approaches -2 is -4.

For the second question, the limit of f(f(x))/5x + 20 as x approaches -4, we can also substitute -4 into the expression:

f(f(-4))/5(-4) + 20

Using the given values in the table, f(-4) is not provided, but we are given f'(-4) as -1. Therefore, we can approximate f(-4) by using the derivative f'(-4).

Using the concept of linear approximation:

f(-4) ≈ f(-2) + f'(-2) * (x - (-2))

≈ -2 + 7 * (-4 - (-2))

≈ -2 + 7 * (-4 + 2)

≈ -2 + 7 * (-2)

≈ -2 + (-14)

≈ -16

Now, we substitute this approximate value back into the expression:

f(f(-4))/5(-4) + 20

= f(-16)/-20 + 20

Using the given values in the table:

= -2/-20 + 20

= 1/10 + 20

= 1/10 + 200/10

= (1 + 200)/10

= 201/10

Therefore, the limit of f(f(x))/5x + 20 as x approaches -4 is 201/10.