# What is the maximum value that the graph of f(x)=4x^2-24x+1?

May 12, 2016

There is no maximum for this function. Just minimum (3,35), the function is convex.

#### Explanation:

Given $f \left(x\right) = 4 {x}^{2} - 24 x + 1$

As $x \rightarrow \pm \infty \textcolor{w h i t e}{\text{XXXX}} f \left(x\right) \rightarrow \infty$

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If you wanted to find the minimum, this could be done by taking the derivative and noting that the derivative will be equal to zero at extrema.

However, this may just be a trick question to see if you will answer with the minimum value without considering that the extreme point is a minimum and not the maximum.

Further details The vertex s given by:

$x = - \frac{b}{2 a} = 3$
$y = - \frac{D}{4 a} = - 35$, where $D = {b}^{2} - 4 a c$

$f ' \left(x\right) = 8 x - 24$, equals to zero: $x = 3$
To find $y$, just replace in the equation: $y = 35$.
$f ' ' \left(x\right) = 8$, it is always positive, which means that you have a minimum.