Differentiation

Lesson

Many polynomials have some or multiple maximums or minimums and usually these are referred to as Maximum Turning Points (the point where the gradient turns from positive to negative), or Minimum Turning Points (the point where the gradient turns from negative to positive).

Turning points can also be

- global: which means that is the absolute maximum or minimum value that the whole function takes is at that point. A quadratic is an example of a function that has a global maximum or minimum.
- or local: which means that the turning point is maximum or minimum in that general area, but that the function may have a larger/smaller value elsewhere. A cubic is an example of a function that can have a local maximum/minimum turning point, but go on elsewhere to have larger/smaller values.

Quadratics have 1 turning point, Cubics can have up to 2, Quartics can have up to 3 ect...

We can identify the location of the turning point by finding the places where the derivative is equal to zero, but this alone is not enough. It just identifies flat spots in the curve, we must also test the points to check if they are turning points.

We do this by testing points either side of where $y'=0$`y`′=0 and looking at the sign of the gradient.

Determine any maximum or minimum values of the function $y=\frac{x^3}{3}-\frac{x^2}{2}-6x+4$`y`=`x`33−`x`22−6`x`+4

The first thing we need is the derivative. $y'=x^2-x-6$`y`′=`x`2−`x`−6

Then we need to solve $y'=0$`y`′=0.

By factorising $y'$`y`′, I can see $y'=(x-3)(x+2)$`y`′=(`x`−3)(`x`+2) which means that the roots of $y'$`y`′ are at $x=3$`x`=3 and $x=-2$`x`=−2.

This means that at $x=3$`x`=3, we have a turning point. Let's check what type.

We will determine the sign of the derivative before and after the value of $x=3$`x`=3, by substituting into $y'$`y`′.

Note - we don't need to know the actual answer, we only need to know if the gradient is positive or negative but you can work out the actual answer if you find that easier.

$y'(0)$`y`′(0). I'm choosing zero because it's an easy one to substitute in. $y'(0)=-6$`y`′(0)=−6, which is negative. So it has a negative gradient here.

$y'(4)$`y`′(4). There are two factors in $y'$`y`′, and when the value of $x$`x` is $4$4 both of these is positive. $+\times+=+$+×+=+. So this is positive gradient here.

You can see from my table that the shape shows us exactly what type of turning point it is. This one at $x=3$`x`=3 is a minimum turning point, (a local one)

Let's check the other one at $x=-2$`x`=−2.

$y'(0)$`y`′(0) we already worked out from before

$y'(-5)$`y`′(−5). The two factors are $(x-3)$(`x`−3) and $(x+2)$(`x`+2) and when $x=-5$`x`=−5, both of these factors are negative. $-\times-=+$−×−=+

So at $x=-2$`x`=−2, we have a maximum turning point (a local one)

Look at the bottom row of this table, it's basically drawn the shape of the cubic nicely for us. Now we put it together on a plane.

The red points are our turning points, the green points I have marked on for you to see how the slope of the curve at these points matches the graph.

So to determine maximum and minimum values

- find $f'(x)$
`f`′(`x`) - set $f'(x)=0$
`f`′(`x`)=0 - test points either side of the points where $f'(x)=0$
`f`′(`x`)=0 and then classify

Consider the parabola with equation $y=x^2-4x+6$`y`=`x`2−4`x`+6.

The vertex of a parabola is located where the derivative is $0$0. If we set the derivative of the given parabola to $0$0, we get $2x-4=0$2

`x`−4=0.Solve this equation to find the $x$

`x`-coordinate of the vertex.Find the $x$

`x`-coordinate of the vertex using the formula $x=-\frac{b}{2a}$`x`=−`b`2`a`.Find the $y$

`y`-coordinate of the vertex.Which of the following statements is true?

The gradient of the tangent to the parabola is positive at $\left(2,2\right)$(2,2).

AThere is a turning point at $\left(2,2\right)$(2,2).

BThe gradient of the tangent to the parabola is negative at $\left(2,2\right)$(2,2).

CThe gradient of the tangent to the parabola is positive at $\left(2,2\right)$(2,2).

AThere is a turning point at $\left(2,2\right)$(2,2).

BThe gradient of the tangent to the parabola is negative at $\left(2,2\right)$(2,2).

C

Consider the function $f\left(x\right)=3x^2-54x+241$`f`(`x`)=3`x`2−54`x`+241

Determine an expression for the derivative of $f\left(x\right)$

`f`(`x`).Hence solve for the $x$

`x`-coordinate of the stationary point.Complete the following table of values.

$x$ `x`$8$8 $9$9 $10$10 $f'\left(x\right)$ `f`′(`x`)$\editable{}$ $\editable{}$ $\editable{}$ Which of these statements is correct?

$\left(9,f\left(9\right)\right)$(9,

`f`(9)) is a minimum turning point.A$\left(9,f\left(9\right)\right)$(9,

`f`(9)) is a maximum turning point.B$\left(9,f\left(9\right)\right)$(9,

`f`(9)) is a minimum turning point.A$\left(9,f\left(9\right)\right)$(9,

`f`(9)) is a maximum turning point.B

Consider the function $f\left(x\right)=\left(x^2-9\right)^2+4$`f`(`x`)=(`x`2−9)2+4.

Expand and simplify the expression for $f\left(x\right)$

`f`(`x`).Determine an expression for the derivative of $f\left(x\right)$

`f`(`x`).Hence locate the $x$

`x`value of the stationary point.Complete the following table of values.

$x$ `x`$-4$−4 $-3$−3 $-2$−2 $f'\left(x\right)$ `f`′(`x`)$\editable{}$ $\editable{}$ $\editable{}$ $x$ `x`$-1$−1 $0$0 $1$1 $f'\left(x\right)$ `f`′(`x`)$\editable{}$ $\editable{}$ $\editable{}$ $x$ `x`$2$2 $3$3 $4$4 $f'\left(x\right)$ `f`′(`x`)$\editable{}$ $\editable{}$ $\editable{}$ Which of these statements is correct?

$\left(-3,f\left(-3\right)\right)$(−3,

`f`(−3)) is a maximum turning point.A$\left(-3,f\left(-3\right)\right)$(−3,

`f`(−3)) is a minimum turning point.B$\left(-3,f\left(-3\right)\right)$(−3,

`f`(−3)) is a maximum turning point.A$\left(-3,f\left(-3\right)\right)$(−3,

`f`(−3)) is a minimum turning point.BSelect the correct statement.

$\left(0,f\left(0\right)\right)$(0,