Show that there is no value of x for which f(x) and g(x) have tangents with same slop

[JSmith]

Show that there is no value of x for which

and g(x) = -x⁵ have tangents with the same slope.
So I calculated the derivatives of both:

Now how would I proceed? Do I set the derivatives equal to eachother??

 

[pka]

Show that there is no value of x for which

and g(x) = -x⁵ have tangents with the same slope.
So I calculated the derivatives of both:

Now how would I proceed? Do I set the derivatives equal to eachother??

\(\displaystyle f'(x)=\dfrac{1}{x^2}~\&~g'(x)=-5x^4\) if \(\displaystyle \forall a\ne 0\) then \(\displaystyle f'(a)>0\) and \(\displaystyle \forall b\) then \(\displaystyle g'(b)\le 0\).

 

[JSmith]

g'(x)=-5x^4, does it not??

Can you explain to me what the statement you made means?

 

[DrPhil]

Show that there is no value of x for which

and g(x) = -x⁵ have tangents with the same slope.
Now how would I proceed? Do I set the derivatives equal to eachother??

Yes. Set f'(x) = g'(x) and solve for x

\(\displaystyle \dfrac{1}{x^2} = -5 x^4 \)

Note that x may not be 0 because neither f(0) nor f'(0) exists. Then

\(\displaystyle x^6 = - \dfrac{1}{5} < 0 \)

But there is no real 6th root of a negative number; therefore x does not exist.

 

[JSmith]

Is this correct???

 

[pka]

Is this correct???

None of that is necessary. All you need to write is that the slope of \(\displaystyle f\) is positive where it is defined and the slope of \(\displaystyle g\) is everywhere not positive. Therefore, the two functions can never have the same slope.

Never over complicate an answer. It does not show that you understand if all you can do is a process.

 

[DrPhil]

After you got the expression over the common denominator of x^2,
the logic for the next step is that the numerator must be 0.
The numerator should have x^6, not x^4 - but the result is the same.

However the reasoning used by pka is simpler and quicker.

 

[JSmith]

Ok, thanks so much for the input! Take care.