1. Basic Calculus HELP!!!!!!!! (:-)

Find the gradient of the tangeant to the curve:

$\displaystyle f(x) = x^3/2 - 5x^2$ at $\displaystyle x = 2$

Find the coordinates of the points at the curve:

$\displaystyle f(x) = (3x - 2)^2$ at which the tangeant is parralel to the line y - 6x = 8.

Find the coordinates of the points on the curve $\displaystyle y = x^3 - x^2 + 5x - 3$ at which the tangeant is perpindicular to $\displaystyle x + 12y = 38$

Find the equation of the curve $\displaystyle y = f(x)$ given $\displaystyle f'(x) = 4x^2 - x + 1$ and $\displaystyle f(0) = 3.$

2. Originally Posted by mibamars
Find the equation of the curve $\displaystyle y = f(x)$ given $\displaystyle f'(x) = 4x^2 - x + 1$ and $\displaystyle f(0) = 3.$

$\displaystyle f'(x) = 4x^2 - x + 1$ and $\displaystyle f(0) = 3.$
$\displaystyle f(x) = (4/3)x^3 - (1/2)x^2 + x + c$
$\displaystyle f(0) = 3 = (4/3)0^3 - (1/2)0^2 + 0 + c$
Therefore, $\displaystyle c = 3$
$\displaystyle f(x) = (4/3)x^3 - (1/2)x^2 + x + 3$

3. Originally Posted by mibamars
(1) Find the gradient of the tangeant to the curve:

$\displaystyle f(x) = x^3/2 - 5x^2$ at $\displaystyle x = 2$

(2) Find the coordinates of the points at the curve:

$\displaystyle f(x) = (3x - 2)^2$ at which the tangeant is parralel to the line y - 6x = 8.

(3) Find the coordinates of the points on the curve $\displaystyle y = x^3 - x^2 + 5x - 3$ at which the tangeant is perpindicular to $\displaystyle x + 12y = 38$

(4) Find the equation of the curve $\displaystyle y = f(x)$ given $\displaystyle f'(x) = 4x^2 - x + 1$ and $\displaystyle f(0) = 3.$
Hello,

to (1):

Calculate the first derivative of f: $\displaystyle f'(x) = \frac32 x^2 - 10x$ and plug in x = 2. That means you are calculating $\displaystyle f'(2) = -14$

to (2):
Rearrange the equation of the given line to: $\displaystyle y = 6x+8$. The slope of the line is m = 6. If the tangent is parallel to this line it's slope must be 6 too. That means you are looking for those values of x that f'(x) = 6.

$\displaystyle f'(x) = 2(3x-2) \cdot 3 = 18x - 12$

$\displaystyle 18x-12 = 6~\iff~ x = 1$ . Plug in this value into the equation of the function. You'll get the coordinates of the tangent point as T(1, 1)

to (3):

Rearrange the equation of the line $\displaystyle x + 12y = 38~\iff~ y = -\frac1{12} x + \frac{19}6$. Therefore the perpendicular direction has the slope m = 12. Thus you are looking for those values of x where f'(x) = 12:

$\displaystyle f'(x) = 3x^2-2x+5$

$\displaystyle 3x^2-2x+5 = 12~\iff~x = \frac13 \pm \frac13 \cdot \sqrt{22}$

You have to plug in these 2 values for x into the original equation of the function to get the y-coordinate of the tangent point.