# Thread: tangent to the function #3

1. ## tangent 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

2. Originally Posted by skeske1234
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?

find the derivative of y and set it equal to the slope of the given line, then solve for x.

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

find the derivative of y and set it equal to 5, then solve for x. don't forget that the question asks for a point on the original curve.
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3. Originally Posted by skeske1234
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
HELLO:There is the solution
1 ) let : $f(x) = 3x^{ - \frac{1}{3}}
$

The dirivate is :
$\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 :

$x + 16y + 3 = 0 \Leftrightarrow y = - \frac{1}{{16}}x - \frac{3}{{16}}$
The tangent is parallel for this line the are same slope $- \frac{1}{{16}}
$

solve : $f'(x) = - \frac{1}{{16}}$
$- \frac{1}{{x^{\frac{4}{3}} }} = - \frac{1}{{16}} \Leftrightarrow x^{\frac{4}{3}} = 16$
I"have :
$\left( {x^{\frac{1}{3}} } \right)^4 = 2^4 \Leftrightarrow x^{\frac{1}{3}} = 2 \Leftrightarrow x = 2^3$
CALCULATE :
$f(2^3 ) = 3\left( {2^3 } \right)^{ - \frac{1}{3}} = \frac{3}{2}$
The point is :
$A\left( {8,\frac{3}{2}} \right)$
LOOK HERE THE GRAPHE

4. Originally Posted by skeske1234
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
Hello same solution in first question

$\begin{array}{l}
g(x) = - x^2 + 3x + 4 \\
g'(x) = - 2x + 3 \\
\end{array}$

Solve this equation :

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

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

the point is

$A( - 1,0)$
LOOK HERE THE GRAPHE