Since its an approximation you need to have the
, as
)
and
+hf'(x))
are not equal (give that
)..
So I'd have something like this.
 & \approx f\left(x\right)+hf'\left(x\right)<br />\\ \text{Let }f\left(x\right) & =x^{\frac{1}{3}}<br />\\ f'\left(x\right) & =\frac{1}{3}x^{-\frac{2}{3}}<br />\\ x=8,\: h=0.06<br />\\ f\left(8+0.06\right) & \approx f\left(8\right)+0.06f'\left(8\right)<br />\\ & =8^{\frac{1}{3}}+0.06\times\frac{1}{3}\left(8^{\frac{1}{3}}\right)^{-2}<br />\\ & =2+0.02\times\frac{1}{4}<br />\\ & =2+0.005<br />\\ & =2.005<br />\\ \implies\sqrt[3]{8.06} & \approx2.005<br />\end{alignedat})
Normally I'd put it as a fraction, but I guess since it's an approximation I guess you can turn it into a decimal, the exaimer's report has both, just make sure at the end when you state
that you use the approximate sign.
For part b, since
)
is decreasing (and positive) at the point that we're approximating our curve with, if we draw a tangent to the curve at
, this tangent line will be above the curve of
)
. This tangent is effectively (given that
is small) what we're using to approximate the value of the curve at our new point, so since the curve is lower than our tangent/approximation line, our approximate value will be greater than the actual value.
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