1. Let f be the function that is defined for all real numbers x and that has the following properties.

(i) f''(x)=24x-18
(ii) f'(1)=-6
(iii) f'(2)=0
a. Find each x such that the line tangent to the graph of f at (x, f (x)) is horizontal
b. Write an expression for f (x)

a. To find the x-values such that the line tangent to the graph of f at (x, f(x)) is horizontal, we need to find the critical points of the function. These occur when the derivative f'(x) equals zero.

Given that f'(x) = 24x - 18, we can set it equal to zero and solve for x:

24x - 18 = 0
24x = 18
x = 18/24
x = 3/4

Therefore, the line tangent to the graph of f at (3/4, f(3/4)) is horizontal.

b. To write an expression for f(x), we integrate f''(x) = 24x - 18 to find f'(x):

∫(f''(x)dx) = ∫(24x - 18)dx
f'(x) = 12x^2 - 18x + C1

Applying condition (ii) f'(2) = 0:

0 = 12(2)^2 - 18(2) + C1
0 = 48 - 36 + C1
C1 = -12

Plugging the value of C1 back into f'(x):

f'(x) = 12x^2 - 18x - 12

Now, we integrate f'(x) to find f(x):

∫(f'(x)dx) = ∫(12x^2 - 18x - 12)dx
f(x) = 4x^3 - 9x^2 - 12x + C2

Since we don't have any conditions for the constant term, we can leave it as C2.

Hence, the expression for f(x) is:
f(x) = 4x^3 - 9x^2 - 12x + C2

a. To find the x-values such that the tangent line to the graph of f at (x, f(x)) is horizontal, we need to find where the derivative of f is equal to zero. Given that f'(x) is the first derivative of f, we can use the information given in property (ii) to determine that f'(2) = 0.

So, one x-value where the tangent line is horizontal is x = 2.

b. To find an expression for f(x), we need to integrate the second derivative of f. Given that f''(x) = 24x - 18, we can integrate it to get the first derivative, f'(x):

∫ f''(x) dx = ∫ (24x - 18) dx
f'(x) = 12x^2 - 18x + C1

Next, we can use the information given in property (i) to determine f'(1) = -6. Plugging this into the expression for f'(x), we get:

-6 = 12(1)^2 - 18(1) + C1
-6 = 12 - 18 + C1
C1 = -6 - 12 + 18
C1 = 0

Now, integrating f'(x), we get f(x):

∫ f'(x) dx = ∫ (12x^2 - 18x + 0) dx
f(x) = 4x^3 - 9x^2 + C2

To find the expression for f(x), we need to determine the value of C2. Using the information in property (iii), we have f'(2) = 0:

0 = 4(2)^3 - 9(2)^2 + C2
0 = 32 - 36 + C2
C2 = 36 - 32
C2 = 4

So, the expression for f(x) is:

f(x) = 4x^3 - 9x^2 + 4

To answer these questions, we can use integration to find the expression for f(x) by integrating f''(x) and then using the initial conditions to find the constant of integration. Then, we can differentiate f(x) to find f'(x) and solve for x when f'(x) equals zero to find the values where the tangent line to the graph of f(x) is horizontal.

a. To find the x-values where the tangent line to the graph of f(x) is horizontal, we need to solve the equation f'(x) = 0.

Given that f'(x) = 24x - 18, we set this equal to zero and solve for x:

24x - 18 = 0
24x = 18
x = 18/24
x = 3/4

So, the x-value where the tangent line to the graph of f(x) is horizontal is x = 3/4.

b. To find the expression for f(x), let's integrate f''(x) = 24x - 18.

∫ f''(x) dx = ∫ (24x - 18) dx

Integrating the right side with respect to x, we get:

f'(x) = 12x^2 - 18x + C1

Now, we'll use the given initial condition f'(1) = -6 to find the value of C1:

f'(1) = 12(1)^2 - 18(1) + C1
-6 = 12 - 18 + C1
-6 = -6 + C1
C1 = 0

So, the expression for f'(x) becomes:

f'(x) = 12x^2 - 18x

Now, let's integrate f'(x) to find f(x):

∫ f'(x) dx = ∫ (12x^2 - 18x) dx

Integrating the right side with respect to x, we get:

f(x) = 4x^3 - 9x^2 + C2

Finally, we'll use the given initial condition f'(2) = 0 to find the value of C2:

f(2) = 4(2)^3 - 9(2)^2 + C2
0 = 32 - 36 + C2
0 = -4 + C2
C2 = 4

So, the expression for f(x) becomes:

f(x) = 4x^3 - 9x^2 + 4

f'(x) = 12x^2 - 18x + c

for f'(1) = -6
12 - 18 + c = -6
c = 0

so f'(x) = 12x^2 - 18x

Then you say f'(2) = 0 , which does not verify with what I have so far.
Did you mean f(2) = 0 ?
I will assume you meant that.
f(x) = 4x^3 - 9x^2 + k
given f(2) = 0
32 - 36 + k = 0
k = 4

f(x) = 4x^3 - 9x^2 + 4

leaving up to you: .....

a) .......
tell me what you would do.

f''(x) = 24x - 18

f'(x) = 12x^2 - 18x + C
Since f'(1) = -6, then
-6 = 12(1)^2 - 18(1) + C
-6 = 12 - 18 + C
-6 = -6 + C
C = 0

So f'(x) = 12x^2 - 18x

f(x) = 4x^3 - 9x^2 + C
Since f(2) = 0, then
0 = 4(2)^3 - 9(2)^2 + C
0 = 4(8) - 9(4) + C
0 = 32 - 36 + C
0 = -4 + C
4 = C

f(x) = 4x^3 - 9x^2 + 4

For Part A, you know that f'(x) = 12x^2 - 18x
f'(x) = 0
12x^2 - 18x = 0
6x(2x-3) = 0
x = 0 or x = 3/2
Just evaluate f(0) and f(3/2) to find the y-coordinates.