hectictar

average rate of change from  f(a)  to  f(b)   =   slope between the points  (a, f(a))  and  (b, f(b))

average rate of change from  f(a)  to  f(b)   =   \(\frac{f(b)-f(a)}{b-a}\)

Let's find  a  using the information that  f(a) = 16

\(f(a)\,=\,\sqrt{5a+7}\\~\\ 16=\sqrt{5a+7}\\~\\ 16^2\,=\,5a+7\\~\\ 256\,=\,5a+7\\~\\ 249\,=\,5a\\~\\ 49.8\,=\,a\)

Let's find  b  using the information that  f(b)  =  25

\(f(b)\,=\,\sqrt{5b+7}\\~\\ 25=\sqrt{5b+7}\\~\\ 625\,=\,5b+7\\~\\ 618\,=\,5b\\~\\ 123.6\,=\,b\)

average rate of change from  f(a)  to  f(b)   =   \(\frac{f(b)-f(a)}{b-a}\)

average rate of change from  f(a)  to  f(b)   =   \(\frac{25-16}{123.6-49.8}\)

average rate of change from  f(a)  to  f(b)   =   \(\frac{9}{73.8}\)

average rate of change from  f(a)  to  f(b)   =   \(\frac{5}{41}\)

(x + 1)(2x + 3)(3x + 4)  =  (6x² + 5)(x - 3)

                                                                                            Multiply  (x + 1)(2x + 3)  and  (6x² + 5)(x - 3)

(2x² + 5x + 3 )(3x + 4)  =  6x³ - 18x² + 5x - 15

                                                                                            Multiply  (2x² + 5x + 3 )(3x + 4)

6x³ + 8x² + 15x² + 20x + 9x + 12  =  6x³ - 18x² + 5x - 15

                                                                                            Combine like terms on the left side.

6x³ + 23x² + 29x + 12  =  6x³ - 18x² + 5x - 15

                                                                            Subtract  6x³  from both sides of the equation.

23x² + 29x + 12  =  -18x² + 5x - 15

                                                                            Add  18x²  to both sides of the equation.

41x² + 29x + 12  =  5x - 15

                                                                            Subtract  5x  from both sides.

41x² + 24x + 12  =  -15

                                                                            Add  15  to both sides.

41x² + 24x + 27  =  0

                                                                            The GCF of  41, 24, and 27  is  1

0  =  41x² + 24x + 27

Now it is in the form   0  =  Ax² + Bx + C  where  A, B, and C are positive integers with a GCF of 1.

A + B + C   =   41 + 24 + 27   =   92

\(\sqrt{25}\)  equals only  5

\(\pm\sqrt{25}\)   equals  ± 5

Hmmmm..I'm still confused....are you talking about

\(\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\frac1k\)   

Guest, how can a sum of positive numbers ever be not positive??

(ii)  False

As a counterexample,

Let     f(x)  =  x - 4     and     a  =  0     and     b  =  6

\(\int_{a}^{b}f(x)\,dx\,=\,\int_{0}^{6}(x-4)\,dx\,=\,(\frac{x^2}{2}-4x)\Big|_{0}^{6}\,=\,-6\,<\,0\)

But....  5  is an element of the interval  [0, 6]  and

\(f(5)\,=\,5-4\,=\,1\nless0\)

The integral is negative because the triangle under the x-axis is bigger than the triangle above the x-axis,

but not all the values of  f(x)  are negative for  x  in the interval  [0, 6] .

\(y\,=\,-3^x-2\)

When     \(x=-2\) ,     \(y\,=\,-3^{-2}-2\,=\,-\frac19-2\,=\,-\frac19-\frac{18}{9}\,=\,-\frac{19}{9}\)

When     \(x=-1\) ,     \(y\,=\,-3^{-1}-2\,=\,-\frac19-2\,=\,-\frac13-\frac{6}{3}\,=\,-\frac{7}{3}\)

When     \(x=\phantom{-}0\) ,     \(y\,=\,-3^0-2\,=\,-1-2\,=\,-3\)

When     \(x=\phantom{-}1\) ,     \(y\,=\,-3^1-2\,=\,-3-2\,=\,-5\)

Can you find what  y  equals when  x  is  2 ?

Multiply each term in the first set of parenthesees by each term in the second set of parenthesees and add the products together. It's "first times first plus first times second plus second times first plus second times second"

(a + b)(c + d)   =   ac  +  ad  +  bc  +  bd

So.....

(8x² + -9)(2x + 4)   =   . . .?

Do you see what to do next? If you still don't understand please say so