NZ Level 7 (NZC) Level 2 (NCEA)
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Evaluate Derivative at a point

Having found the derivative function, you can evaluate the derivative at a point by substitution in any value of $x$x.

For the following derivative function $f'(x)=3x^2-4x+1$f(x)=3x24x+1 the values of the derivative at the points on the curve where $x=-1,0$x=1,0 and $1$1

If the original function was $f(x)=x^3-2x^2+x+7$f(x)=x32x2+x+7.  Then it's graph would look like this. 

The points we are interested in are at $x=-1,0$x=1,0 and $1$1.  Marked here on the graphs with their tangents.

Now we know that the derivative function is $f'(x)=3x^2-4x+1$f(x)=3x24x+1, so the derivative at $x=-1$x=1 is $f'(-1)$f(1), the derivative at $x=0$x=0 is $f'(0)$f(0) and the derivative at $x=1$x=1, is $f'(1)$f(1).

$f'(-1)=3(-1)^2-4(-1)+1=3+4+1=8$f(1)=3(1)24(1)+1=3+4+1=8   This is pretty steep, and the graph above confirms this.

$f(0)=3(0)^2-4(0)+1=1$f(0)=3(0)24(0)+1=1.  This looks reasonable on the graph above

$f(1)=3(1)^2-4(1)+1=3-4+1=0$f(1)=3(1)24(1)+1=34+1=0.  And we can tell the horizontal tangent on the graph above also. 


All we need to be able to evaluate the derivative, is the $x$x coordinate of the point and the gradient function.  


Question 1

Simplify $f'\left(x\right)=\frac{x^2\left(7x^2\right)-\left(x^3-5\right)\left(4x\right)}{\left(x^2\right)^2}$f(x)=x2(7x2)(x35)(4x)(x2)2.

Question 2

Find the gradient of $f\left(x\right)=16x^{-3}$f(x)=16x3 at $x=2$x=2.

Denote this gradient by $f'\left(2\right)$f(2).

Question 3

By considering the graph of $f\left(x\right)=-6$f(x)=6, find $f'\left(4\right)$f(4).



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