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
Let's piece it all together through an example.
Sketch the function $y=\frac{1}{4}(x^3-6x^2+3x+10)$y=14(x3−6x2+3x+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.
$y$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(x3−6x2+3x+10)
so
$y'=\frac{1}{4}(3x^2-12x+3)$y′=14(3x2−12x+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(3x2−12x+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
$x^2-7x+10=(x-5)(x-2)$x2−7x+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
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 |
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 |