1. Find the values of x so that the tangent to the function y=3/(x^1/3) is parallel to the line x+16y+3=0?

2. Find the point on the parabola y=-x^2+3x+4 where the slope of the tangent is 5

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- Aug 4th 2009, 04:50 PMskeske1234tangent to the function #3
1. Find the values of x so that the tangent to the function y=3/(x^1/3) is parallel to the line x+16y+3=0?

2. Find the point on the parabola y=-x^2+3x+4 where the slope of the tangent is 5 - Aug 4th 2009, 06:23 PMskeeter
- Aug 7th 2009, 12:36 PMdhiab
HELLO:There is the solution

1 ) let : $\displaystyle f(x) = 3x^{ - \frac{1}{3}}

$

The dirivate is :

$\displaystyle \begin{array}{l}

f'(x) = 3\left( { - \frac{1}{3}} \right)x^{ - \frac{1}{3} - 1} \\

f'(x) = - \frac{1}{{x^{\frac{4}{3}} }} \\

\end{array}

$

But :

$\displaystyle x + 16y + 3 = 0 \Leftrightarrow y = - \frac{1}{{16}}x - \frac{3}{{16}}$

The tangent is parallel for this line the are same slope $\displaystyle - \frac{1}{{16}}

$

solve : $\displaystyle f'(x) = - \frac{1}{{16}}$

$\displaystyle - \frac{1}{{x^{\frac{4}{3}} }} = - \frac{1}{{16}} \Leftrightarrow x^{\frac{4}{3}} = 16$

I"have :

$\displaystyle \left( {x^{\frac{1}{3}} } \right)^4 = 2^4 \Leftrightarrow x^{\frac{1}{3}} = 2 \Leftrightarrow x = 2^3 $

CALCULATE :

$\displaystyle f(2^3 ) = 3\left( {2^3 } \right)^{ - \frac{1}{3}} = \frac{3}{2}$

The point is :

$\displaystyle A\left( {8,\frac{3}{2}} \right)$

LOOK HERE THE GRAPHE - Aug 7th 2009, 12:56 PMdhiab
Hello same solution in first question

$\displaystyle \begin{array}{l}

g(x) = - x^2 + 3x + 4 \\

g'(x) = - 2x + 3 \\

\end{array}$

Solve this equation :

$\displaystyle g'(x) = 5 \Leftrightarrow - 2x + 3 = 5 \Leftrightarrow x = 5$

Calculate :

$\displaystyle g(5) = - \left( { - 1} \right)^2 + 3\left( { - 1} \right) + 4 = 0$

the point is

$\displaystyle A( - 1,0)$

LOOK HERE THE GRAPHE