# How do you find the max or minimum of f(x)=x-2x^2-1?

May 17, 2017

This parabola has a maximum point at vertex: $\left(\frac{1}{4} , - \frac{7}{8}\right)$

#### Explanation:

When you have a parabola in standard form

$f \left(x\right) = a {x}^{2} + b x + c$

Here are some rules

• If $a > 0$, then vertex is a minimum
• If $a < 0$, then vertex is a maximum
• Vertex is given by the following formula

Vertex: $\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$

Step 1. Rewrite the equation in standard form
$f \left(x\right) = - 2 {x}^{2} + x - 1$
$a = - 2$
$b = 1$
$c = - 1$

Step 2. Determine maximum or minimum
$a = - 2 < 0$, so the vertex represents a maxium

Step 3. Plug into the vertex formula
$x$-value of the vertex: $x = \frac{- 1}{2 \left(- 2\right)} = \frac{- 1}{-} 4 = \frac{1}{4}$

$y$-value of the vertex: $f \left(\frac{1}{4}\right) = - 2 {\left(\frac{1}{4}\right)}^{2} + \frac{1}{4} - 1 = - \frac{7}{8}$

Vertex: $\left(\frac{1}{4} , - \frac{7}{8}\right)$

graph{-2x^2+x-1 [-2, 2, -5, 0.1]}

May 17, 2017

refer below

#### Explanation:

The given equation is Quadratic, which means it only has 1 maximum or 1 minimum point.

1. Re-arranging the equation, $f \left(x\right) = - 2 {x}^{2} + x - 1$

Since the coefficient of the term ${x}^{2}$ is a negative, the graph of the curve has a maximum point. Conversely, if it is a positive, it has a minimum point.

graph{-2x^2+x-1 [-4.896, 5.104, -4.66, 0.34]}

The curve is shown in this graph.

2. At the turning point, the gradient of the function is zero.

This means $\textcolor{red}{f ' \left(x\right) = 0}$

Differentiating the function,

$f ' \left(x\right) = - 4 x + 1 = \textcolor{red}{0}$

$- 4 x = - 1$

$\textcolor{b l u e}{x = \frac{1}{4}}$

3. Most questions require you to prove/determine the turning point of the function is a maximum/minimum point.

To prove that its turning point is a maximum, you have to use the Second Derivative Test.

A maximum point will have $\textcolor{red}{f ' ' \left(x\right) < 0}$
(a minimum point will have f''(x)>0)

Differentiating f'(x),

$f ' ' \left(x\right) = - 4$

Since $f ' ' \left(x\right) < 0$, the turning point is a maximum

4. To find the coordinates of the maximum, substitute the $\textcolor{b l u e}{x}$ value into f(x).

when $\textcolor{b l u e}{x = \frac{1}{4}}$,

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

$y = - \frac{7}{8}$

5. Coordinates of the maximum is $\left(\frac{1}{4} , - \frac{7}{8}\right)$.