Question

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

Answer 1

This parabola has a maximum point at vertex: `(1/4, -7/8)`

Explanation:

When you have a parabola in standard form

`f(x)=ax^2+bx+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: `(-b/(2a), f(-b/(2a)))`

Step 1. Rewrite the equation in standard form
`f(x)=-2x^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=(-1)/(2(-2))=(-1)/-4=1/4`

`y`-value of the vertex: `f(1/4)=-2(1/4)^2+1/4-1=-7/8`

Vertex: `(1/4, -7/8)`

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

Answer 2

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(x)=-2x^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 `color(red)(f'(x)=0)`

Differentiating the function,

`f'(x)= -4x+1 = color(red)0`

`-4x=-1`

`color(blue)(x=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 `color(red)(f''(x)<0)`
(a minimum point will have f''(x)>0)

Differentiating f'(x),

`f''(x)=-4`

Since `f''(x)<0`, the turning point is a maximum

4. To find the coordinates of the maximum, substitute the `color(blue)x` value into f(x).

when `color(blue)(x=1/4)`,

`f(x) = y = -2(1/4)^2+1/4-1`

`y=-7/8`

5. Coordinates of the maximum is `(1/4,-7/8)`.