# advanced functions intro calc questions

• Feb 22nd 2009, 10:22 PM
jamman790
ok so im very new to calculus and just started learning limits...were exiting logs and entering limits (im a grade 12 student) my teacher has given me 2 question that have got me stumped.. here they are if anyone can help me out pllleaase help:) thanks:)

solve for x

2^x - 2^-x =4

2)determine the equation of both lines that are tangent to the graph of
f( x)=x^2
and pass through point
(1 , -3)
Must use
lim F( x ) - F ( a ) / x - a
x->a
• Feb 23rd 2009, 01:10 AM
earboth
Quote:

Originally Posted by jamman790
ok so im very new to calculus and just started learning limits...were exiting logs and entering limits (im a grade 12 student) my teacher has given me 2 question that have got me stumped.. here they are if anyone can help me out pllleaase help:) thanks:)

1) solve for x

2^x - 2^-x =4

2)determine the equation of both lines that are tangent to the graph of
f( x)=x^2
and pass through point
(1 , -3)
Must use
lim F( x ) - F ( a ) / x - a
x->a

to #1):

$2^x-2^{-x} = 4~\implies~(2^x)^2-1=4\cdot 2^x~\implies~2^{2x}-4 \cdot 2^x-1=0$

This is a quadratic in $2^x$ . Use the substitution $y = 2^x$
Solve for y
Re-substitute to calculate the value(s) of x

to #2)

Let $T(t, t^2)$ denote the tangent point on the graph of f.

The slope of the tangent is $m_t=f'(t) = 2t$

The equation of the tangent in T is:

$y-t^2=2t(x-t)~\implies~\boxed{y=2tx-t^2}$

This tangent has to pass through P(1, -3) that means the coordinates of P have to satisfy the equation of the tangent:

$-3=2t\cdot 1-t^2~\implies~t^2-2t-3=0~\implies~t=3~\vee~t=-1$

You get two tangent points and two tangents:

$T_1(-1,1)$ with tangent $y = -2x-1$

$T_2(3, 9)$ with tangent $y = 6x-9$
• Feb 23rd 2009, 09:56 AM
jamman790
can u please explain for the 1st question you awnsered (x^2 - X^-2 = 4) how you get to the second step or more specifically how u ended up with x^2x - 1? thank you:)
• Feb 23rd 2009, 10:19 AM
HallsofIvy
Quote:

Originally Posted by jamman790
can u please explain for the 1st question you awnsered (x^2 - X^-2 = 4) how you get to the second step or more specifically how u ended up with x^2x - 1? thank you:)

He doesn't. I think you are misreading " $(2^x)^2- 1= 4(2^x)$"
From $2^x- 2^{-x}= 4$, multiply both sides of the equation by $2^x$: $(2^x)(2^x)- (2^x)(2^{-x})= (2^x)(4)$ which is the same as $(2^x)^2- 1= 4(2^x)$. Now let $y= 2^x$ and that equation becomes $y^2- 1= 4y$ or $y^2- 4y- 1= 0$.

Myself, I would have made that substitution to begin with: Let y= $2^x$ in the equation $2^x- 2^{-x}= 4$ so that you have $y- \frac{1}{y}= 4$ and multiply on both sides by y: $y^2- 1= 4y$, which is, of course, exactly the same thing.
• Feb 23rd 2009, 11:00 AM
jamman790
OooooH! Thank you so much:D thanks guys, life saver:)