horizontal tangent

• Feb 28th 2008, 06:51 PM
ArmiAldi
horizontal tangent
7. Find all values of x for the given function where the tangent line is horizontal.

f(x) = root( x^2 + 18x + 86)

9. Find derivative

y = ( 6e^x) / ( 2e^x + 1)
• Feb 28th 2008, 09:57 PM
earboth
Quote:

Originally Posted by ArmiAldi
7. Find all values of x for the given function where the tangent line is horizontal.

f(x) = root( x^2 + 18x + 86)

9. Find derivative

y = ( 6e^x) / ( 2e^x + 1)

$\displaystyle f(x) = \sqrt{ x^2 + 18x + 86} = (x^2+18x+86)^{\frac12}$ . Now calculate the first derivative:

$\displaystyle f'(x)= \frac12 \cdot (x^2+18x+86)^{-\frac12} \cdot (2x+18) = \frac{2x+18}{{\frac12}\cdot \sqrt{x^2+18x+86}}$

f'(x) = 0 if the numerator equals zero and the denominator is not zero:

$\displaystyle 2x+18=0~\implies~x=-9$

$\displaystyle f(x)=\frac{6e^x}{2e^x + 1}$ . Use quotient rule:

$\displaystyle f'(x) = \frac{(2e^x + 1) \cdot 6e^x - 6e^x \cdot 2e^x}{(2e^x+1)^2} = \frac{6e^x}{(2e^x+1)^2}$
• Feb 28th 2008, 10:30 PM
mr fantastic
Quote:

Originally Posted by earboth;111724[snip
$\displaystyle f(x)=\frac{6e^x}{2e^x + 1}$ . Use quotient rule:

$\displaystyle f'(x) = \frac{(2e^x + 1) \cdot 6e^x - 6e^x \cdot 2e^x}{(2e^x+1)^2} = \frac{6e^x}{(2e^x+1)^2}$

Alternatively, note that

$\displaystyle f(x)=\frac{6e^x}{2e^x + 1} = \frac{(6e^x + 3) - 3}{2e^x + 1} = 3 - \frac{3}{2e^x + 1}$.

The derivative of the first term is obviously zero. To get the derivative of the second term, you can use either the chain or quotient rule.