- #1

chwala

Gold Member

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- Homework Statement:
- Draw the tangent to the curve at the point where ##x=1##. Use this tangent to claculate an estimate to the gradient of ##y=2^x## when ##x=1##

- Relevant Equations:
- gradient

Ok this is a question that i am currently marking...the sketch is here;

In my mark scheme i have points ##(1,2)## and ##(3,5)## which can be easily picked from the graph to realize an estimate of ##m=1.5## where ##m## is the gradient ...of course i have given a range i.e ##1.6≥m≥1.2##

Now to my question. hmmmmm

A student picked the points ##(1,2)## and ##(0.9,1.8)## getting ##m=2## ...the difference from actual is quite big...but the points are picked from their straight line...am i missing something here...

Actual gradient using differentiation would be given by;

##\dfrac{dy}{dx}= 2^x\ln 2##

##\dfrac{dy}{dx}[x=1]= 2^1\ln 2=1.386##

Your insight welcome.

In my mark scheme i have points ##(1,2)## and ##(3,5)## which can be easily picked from the graph to realize an estimate of ##m=1.5## where ##m## is the gradient ...of course i have given a range i.e ##1.6≥m≥1.2##

Now to my question. hmmmmm

A student picked the points ##(1,2)## and ##(0.9,1.8)## getting ##m=2## ...the difference from actual is quite big...but the points are picked from their straight line...am i missing something here...

Actual gradient using differentiation would be given by;

##\dfrac{dy}{dx}= 2^x\ln 2##

##\dfrac{dy}{dx}[x=1]= 2^1\ln 2=1.386##

Your insight welcome.

Last edited: