## anonymous 4 years ago 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

1. anonymous

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.

2. anonymous

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

3. anonymous

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].

4. anonymous

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

5. anonymous

thank you so much!!!

6. anonymous

You're welcome.