Thread: What value of x is a tangent parallel to the secant

Good evening folks.

f(x) = 3^x - x^3
For what value of x, is the tangent to the curve parallel to the secant through (0,1) and (3,0)?

Do I find the slop between those 2 points and set it equal to the derivative of the original function?

Originally Posted by alny

Good evening folks.

f(x) = 3^x - x^3
For what value of x, is the tangent to the curve parallel to the secant through (0,1) and (3,0)?

Do I find the slop between those 2 points and set it equal to the derivative of the original function?

Yes that's exactly what you do. Then solve for .

Originally Posted by Prove It

Yes that's exactly what you do. Then solve for .

Wow! Is it really? So if I do that it will turn into a logarithmic equation, correct? And solving that will get me my x?

Originally Posted by alny

Wow! Is it really? So if I do that it will turn into a logarithmic equation, correct? And solving that will get me my x?

Yes.

Just remember though that you will need to use the chain rule to differentiate .

3^x(ln3)-3x^2 = -1/3 ?

Originally Posted by alny

3^x(ln3)-3x^2 = -1/3 ?

Correct. Now solve for x if you can.

Originally Posted by Prove It

Correct. Now solve for x if you can.

This is as far as I got and i lost confidence that i was doing the right thing..

3^x(ln3)-3x^2=-1/3 ... multiplied both sides by 3 to get rid of the fraction

9^x(ln3) - 9x^2 = -1

log(9^x(ln3) + 1) = log(9x^2)

xlog9(ln3) + log(1) = 2log9x

did I royally screw up somewhere?

Continuing on..

xlog9(ln3)-2log9x=log1
x(log9ln(3) -2log9) = log 1
x = log 1 / log9ln(3)-2log9
x = 0 ??????

Originally Posted by alny

log(9^x(ln3) + 1) = log(9x^2)

xlog9(ln3) + log(1) = 2log9x

I don't think you can split a log up like that. Log(a+b) isn't Log(a) + Log(b)

By the way, here are the answers for your eqaution: solve 3^x ln(3) -3x^2 = (-1/3) for x - Wolfram|Alpha