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].
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 <span>[18+/-2√(393)]/24 are the x values of f(x).
( </span>[18+/-2√(393)]/24 , f ( <span>[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)
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