functions

[unborn stone]

i took f(x) as a^x and got a=5 with the given info f(1) but i cant work around th info f'(0)
help pls!

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[fallow ginkgo]

This is called the cauchy exponential functional equations if you want extra details

But you don't need that to solve the question

Let x=y=z/2 so you get f(z)=f(z/2)^2

Take the derivative wrt x

f'(z)=2f(z/2)`+f(z/2)'

Now f'(0)=3 is given and also 0/2=0 so

3=2(3)+f(0) sooo ... Complete this

[unborn stone]

oh i didnt know about them...ill learn about them right now

[unborn stone]

yes i think i will able to solve it now

thanks a lot

[fallow ginkgo]

For f'(1) use the definition of limit you know f(x+h)-f(x)/h one

Then use the original equation in terms of x and y to find f(1+h)= what

[fallow ginkgo]

It won't be required for your course but its just an extra thing

You're not expected to know that to solve this

[unborn stone]

sure i will

[unborn stone]

ohh i think ill have a glance...quite piqued

[fallow ginkgo]

Have fun 😁