Question

Calc Q: let f be the fxn that is defined for all real numbers x and that has the follwing properties... f''(x) =24x-18, f'(-1) =-6, and f(2) =0. A) find each x such hat the line tangent to the graph of f at (x,f(x) ) is horizontal. B) write an expression for f(x) . C) find the average value of f on the interval [1,3].

Answer 1

The solution involves finding where the first derivative of f(x) is zero for a horizontal tangent line, integrating f''(x) to find f(x), and computing the average value of f(x) on the interval [1,3].

Calculus Question: Finding Properties of Function f(x)

A student asked about finding certain properties of a function f(x) that is defined for all real numbers and is governed by second and first derivatives f''(x) and f'(x) along with an initial function value. The function has the following known properties: f''(x) = 24x - 18, f'(-1) = -6, and f(2) = 0.

Part A: Finding the Horizontal Tangents

To find the values of x where the tangent to the graph of f is horizontal, we look for where f'(x) = 0. The second derivative f''(x) gives us the acceleration or the concavity of the graph. For a horizontal tangent, the first derivative which represents velocity or the slope of the tangent line, must be zero.

Part B: Expression for f(x)

To write an expression for f(x), we need to integrate the given second derivative f''(x) to find f'(x), then integrate f'(x) to find f(x). We use the initial values provided to solve for the constants of integration.

Part C: Average Value of f(x) on [1,3]

The average value of f(x) on the interval [1,3] is found using the integral of f(x) over the interval divided by the interval length. This involves integrating f(x) from 1 to 3 and then dividing by 2.

Answer 2

Because we will have to do it eventually, let's find f(x) first.

f''(x) = 24x - 18

First, we integrate f''(x) using an indefinite integral.

f'(x) = ∫ (24x - 18) dx

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

Now, we need to find C by substituting "x" for -1 and setting the equation equal to -6 because f'(-1) = -6

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

Solve for C

C=-26

Now we put that into f'(x).

f'(x) = 12x^2 - 18x -26

Now, we integrate again.

f(x) = ∫(12x^2 - 18x - 26)dx

f(x) = 4x^3 - 9x^2 - 26x + C

Now, we substitute "x" for 2 and set it equal to 0.

f(2) = 4(2)^3 - 9(2)^2 - 26(2) + C = 0

Solve for C

C=38

Now that we have f(x), B has been solved.

Next, we need to find out where the slope of f(x) is equal to 0. Remember that to find slope, we need to find the derivative. We already found the derivative of f(x), so we can use that. The question asks for the places where the slope of f(x) is 0, so we need to set f'(x) equal to 0 and solve for "x"

12x^2 - 18x -26 = 0

I could try to factor this, but I know that it is not possible. We must use the quadratic formula. I cannot reasonable put the quadratic formula into this, so I will only do the part under the radical.

√[18^2-4(12)(-26)]

√(324+1248)

√1572

√(4)*√(393)

2√393

When this is done with the rest of the formula, you get.

[18+/-2√(393)]/24

Thus, the points where the slope of f(x) is equal to zero are where [18+/-2√(393)]/24 are the x values of f(x).

( [18+/-2√(393)]/24 , f ( [18+/-2√(393)]/24 ) )

Now, we have done A and B.

To do C, we must remember the formula for finding the average value of f(x)

This formula is:

[The definite integral of f(x)] / (b-a)

The picture is of this formula.

When we solve for this, we get -13.

I hope I got everything right.