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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

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

  5. anonymous
    • 4 years ago
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    thank you so much!!!

  6. anonymous
    • 4 years ago
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    You're welcome.

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