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
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 : $\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
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