Differentiation

New Zealand

Level 7 - NCEA Level 2

Lesson

Knowing what we know about how a derivative can help us identify key features of the original function, we can now move on to sketching functions from the derivative information.

To sketch a function we need to know

- The general shape the function will have. Knowing the degree of the function will tell us the general shape and end behaviour. The derivative function is always one degree less than the original function.
- The sign of the coefficient of the derivative function will match the sign of the original function. So if the derivative function is a negative quadratic, then the original function is a negative cubic. This tells us the general shape.
- The zeros of the derivative will tell us where the original function has a gradient of zero (horizontal tangent)
- If we check for the sign of the gradient at points either side of the where the gradient is zero we can piece together the general shape of the curve.
- $x$
`x`- and $y$`y`-intercepts can be found by using the original function.

Let's piece it all together through an example.

Sketch the function $y=\frac{1}{4}(x^3-6x^2+3x+10)$`y`=14(`x`3−6`x`2+3`x`+10)

When sketching functions the form the equation is in always tells us something about the shape of the function, but we will need to do some further manipulation to get the whole story.

- The shape is a positive cubic curve, so will look something like this

- The $y$
`y`-intercept occurs at the point where $x=0$`x`=0, so the $y$`y`-intercept is at $y=\frac{1}{4}(0-0+0+10)=\frac{10}{4}=2.5$`y`=14(0−0+0+10)=104=2.5

$y$`y`-intercept $(0,2.5)$(0,2.5)

So far this could look like a lot of cubics.

We need more information to narrow it down.

Let's use the derivative.

$y=\frac{1}{4}(x^3-6x^2+3x+10)$`y`=14(`x`3−6`x`2+3`x`+10)

so

$y'=\frac{1}{4}(3x^2-12x+3)$`y`′=14(3`x`2−12`x`+3)

The roots of the derivative will tell us where the gradient of the curve is $0$0.

So we set $y'=0$`y`′=0 and solve.

$0=\frac{1}{4}(3x^2-12x+3)$0=14(3`x`2−12`x`+3)

$0$0 | $=$= | $\frac{1}{4}(3x^2-12x+3)$14(3x2−12x+3) |

$0$0 | $=$= | $3x^2-12x+3$3x2−12x+3 |

$0$0 | $=$= | $x^2-4x+1$x2−4x+1 |

$x$x |
$=$= | $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$−b±√b2−4ac2a |

$x$x |
$=$= | $\frac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4}}{2}$−(−4)±√(−4)2−42 |

$x$x |
$=$= | $\frac{4\pm\sqrt{16-4}}{2}$4±√16−42 |

$x$x |
$=$= | $\frac{4\pm2\sqrt{3}}{2}$4±2√32 |

$x$x |
$=$= | $\frac{4+2\sqrt{3}}{2},\frac{4-2\sqrt{3}}{2}$4+2√32,4−2√32 |

$x$x |
$=$= | $3.7320508...and0.26794919...$3.7320508...and0.26794919... |

So now we know that at those points, the gradient of the tangent is $0$0.

The full coordinate of these points can be found by substituting into the original function. Let's use $1$1 decimal place for now to keep it a little more simple.

$y(3.7)=\frac{1}{4}(3.7^3-6\times3.7^2+3\times3.7+10)=-2.6$`y`(3.7)=14(3.73−6×3.72+3×3.7+10)=−2.6

So the coordinate is $(3.7,-2.6)$(3.7,−2.6)

$y(0.3)=\frac{1}{4}(0.3^3-6\times0.3^2+3\times0.3+10)=2.6$`y`(0.3)=14(0.33−6×0.32+3×0.3+10)=2.6

So the coordinate is $(0.3,2.6)$(0.3,2.6)

We call these points turning points as the gradient turns from positive to negative and negative to positive at these points.

Now we have the following confirmed information.

The last piece of the puzzle would be the roots of the function, we can find these by solving the initial function.

This is a cubic, so without using technology the way I will solve it will be to first find an identifiable factor, then fully factorise the cubic, and then solve.

Testing for factors. Try

Try $x=0$`x`=0, $y(0)=10$`y`(0)=10

Try $x=1$`x`=1, $y(1)=\frac{1}{4}(1-6+3+10)=2$`y`(1)=14(1−6+3+10)=2

Try $x=-1$`x`=−1, $y(-1)=\frac{1}{4}(-1-6-3+10)=0$`y`(−1)=14(−1−6−3+10)=0

So as $x=-1$`x`=−1 is a root, then $(x+1)$(`x`+1) is a factor.

Use polynomial division

Now we factorise the quadratic

$x^2-7x+10=(x-5)(x-2)$`x`2−7`x`+10=(`x`−5)(`x`−2)

So the fully factorised cubic is $y=\frac{1}{4}(x+1)(x-2)(x-5)$`y`=14(`x`+1)(`x`−2)(`x`−5)

Which gives us the roots of $x=-1$`x`=−1, $x=2$`x`=2 and $x=5$`x`=5.

Now it's basically a dot to dot to construct the function.

Sketch the linear function that satisfies the following information

$f\left(0\right)=1$`f`(0)=1

$f'\left(2\right)=3$`f`′(2)=3

Plot the line on the graph

Loading Graph...

Consider the four functions sketched below.

Which of the sketches match the information for $f\left(x\right)$`f`(`x`) shown in the table? Select all the correct options.

Information about $f\left(x\right)$f(x): |

$f\left(0\right)=5$f(0)=5 |

$f\left(-2\right)=0$f(−2)=0 |

$f'\left(3\right)=0$f′(3)=0 |

$f'\left(x\right)>0$f′(x)>0 for $x<3$x<3 |

- ABCD

Consider the four functions sketched below.

Which of the sketches match the information for $f\left(x\right)$`f`(`x`) shown in the table? Select all the correct options.

Information about $f\left(x\right)$f(x): |

$f'\left(-1\right)=0$f′(−1)=0 |

$f'\left(4\right)=0$f′(4)=0 |

$f'\left(x\right)>0$f′(x)>0 for $x>4$x>4 |

$f'\left(x\right)<0$f′(x)<0 elsewhere |

- ABCD

Sketch the graphs of functions and their gradient functions and describe the relationship between these graphs

Apply calculus methods in solving problems