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Find the values of b such that the function has the given maximum value.

                                              f(x) =  –  x^2  +  bx  –  75

The main objective of this question is to find the maximum or minimum value of the given function.

This question uses the concept of the maximum and minimum value of the function.  The maximum value of the function is the value where the given function touches the graph at its peak value while the minimum value of the function is the value where the function touches the graph at its lowest value.

Expert Answer

We have to find the $b$ value for which the function gives a maximum value of $86$.

The standard form of the equation which gives maximum value is:

\[f(x)\space = \space a(x-h)^2 \space + \space k \]

The given equation is:

\[f(x) \space = \space -x^2 \space\]

\[=\space – \space (x^2 \space – \space bx) \space – \space 75)\]

Now adding the term $\frac{b^2}{4} –  \frac{b^2}{4}$ to the expression results in:

\[= \space – \space (x^2 \space – \space bx \space + \space \frac{b^2}{4} \space – \space \frac{b^2}{4} \space) \space – \space 75 \]

\[= \space – \space (x^2 \space – \space bx \space + \space \frac{b^2}{4}) \space + \space \frac{b^2}{4}  \space – \space 75 \]

\[\space = \space  – \space (x \space – \space \frac{b}{2})^2 \space – \space 75 \space + \space \frac{b^2}{4}\]

Now the equation is in the standard form. The formula is:

\[k \space = \space \frac{b^2}{4} \space –  \space 75\]

Let $k \space=\space25$ to find the value of b.

\[25 \space = \space \frac{b^2}{4} \space –  \space 75\]

\[100 \space = \space \frac{b^2}{4}\]

\[400 \space = \space b^2\]

Taking the squareroot on both side results in:

\[b \space = \space \pm 20\]

Numerical Answer

The given function has a maximum value of  $25$ for b equal to \pm20.

Example

Find the maximum or minimum value of the given function which has a maximum value of $86$.

– $f(x) \space = \space – \space x^2 \space + \space bx \space- \space 14$

The standard form and mathematical representation  of the equation which gives maximum value is:

\[f(x)\space = \space a(x-h)^2 \space + \space k \]

The given equation  for which we have to find the maximum value is:

\[f(x) \space = \space -x^2 \space\]

\[=\space – \space (x^2 \space – \space bx) \space – \space 14)\]

Adding the term $\frac{b^2}{4} –  \frac{b^2}{4}$ to the expression results in:

\[= \space – \space (x^2 \space – \space bx \space + \space \frac{b^2}{4} \space – \space \frac{b^2}{4} \space) \space – \space 14 \]

\[= \space – \space (x^2 \space – \space bx \space + \space \frac{b^2}{4}) \space + \space \frac{b^2}{4}  \space – \space 14 \]

\[\space = \space  – \space (x \space – \space \frac{b}{2})^2 \space – \space 14 \space + \space \frac{b^2}{4}\]

Now the equation is in the standard form. We know the formula as:

\[k \space = \space \frac{b^2}{4} \space –  \space 14\]

Let $k \space=\space 86$ to find the value of b.

\[86 \space = \space \frac{b^2}{4} \space –  \space 14\]

\[100 \space = \space \frac{b^2}{4}\]

Simplifying the above equation results in:

\[400 \space = \space b^2\]

Taking the square root on both sides results in:

\[b \space = \space \pm 20\]

Hence, the maximum value for the given expression is $86$ for b equal to \pm20.

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