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 teh interval [1,3]

i don't need process in details, but just a general idea of the process to use

Answer 1

a. You need to find all points \((x,f(x))\) such that \(f'(x)=0\). We can find f' by integrating f'' once.
\[f'(x)=\int (24x-18)dx=12x^2-18x+c\]
But \(f'(-1)=-6 \implies 12(-1)^2-18(-1)+c=-6 \implies c=-36.\)
Now solve the quadratic equation \(f'(x)=12x^2-18x-36=0\) to find those points.

Answer 2

b. To find \(f(x)\), integrate f' and then use the point f(2)=0 to find the constant (as I did with f').

Answer 3

c. Use the formula
\[f_{\text{avg}}=\frac{1}{b-a}\int_a^bf(x)\]
This formula gives you the average value of a function f(x) on the interval [a,b].

Answer 4

Obviously in your case, a=1, b=3 and f is what you will find from part b.

Answer 5

thank you so much!!!

Answer 6

You're welcome.