1. ## A few questions

Could someone please help me with these couple questions? Instructions on how to do the first two would be great, but for 3 and 4 I know how to do them but I'm not getting the right answers.

1. Determine a quadratic function f(x)=ax²+bx+c whose graphy passes through the point (2,19) and that has a horizontal tangent at (-1,-8).

2. For what values of x do the curves y=(1+x³)² and y=2x6 (6 is an exponent on the x) have the same slope?

3. Differentiate y= (1+√x / x2/3 )³, where 2/3 is an exponent on that x in the denominator.

4. Differentiate y= (√(1-x²) / 1-x ).

Thanks a ton!!!

2. Originally Posted by NAPA55
1. Determine a quadratic function f(x)=ax²+bx+c whose graphy passes through the point (2,19) and that has a horizontal tangent at (-1,-8).
The function passes through (2, 19), so we know that
$19 = 4a + 2b + c$

And the function passes through (-1, -8), so
$-8 = a - b + c$

The derivative of the function is
$f^{\prime} = 2ax + b$

We know that the tangent is horizontal (that the first derivative is equal to 0) at x = -1:
$0 = -2a + b$

Three equations, three unknowns.

-Dan

3. Originally Posted by NAPA55
2. For what values of x do the curves y=(1+x³)² and y=2x6 (6 is an exponent on the x) have the same slope?
The equations have the same slope, so the derivatives are equal for those x values. So find the derivatives, set them equal to each other, and solve for x.

-Dan

4. Originally Posted by NAPA55
3. Differentiate y= (1+√x / x2/3 )³, where 2/3 is an exponent on that x in the denominator.
$y = \left ( \frac{1 + \sqrt{x}}{x^{2/3}} \right ) ^3$

This is an exercise in the chain rule:
$y^{\prime} = \left [ 3 \left ( \frac{1 + \sqrt{x}}{x^{2/3}} \right )^2 \right ] \cdot \left [ \frac{\frac{1}{2 \sqrt{x}} \cdot x^{2/3} - (1 + \sqrt{x} ) \cdot \frac{2}{3}x^{-1/3} }{x^{4/3}} \right ]$

The first set of [ ] is the derivative of the "outer" function, the cube. The second set of [ ] is the derivative of the fraction inside the ( ). Of course, you still need to simplify this.

-Dan

5. Originally Posted by topsquark
$y = \left ( \frac{1 + \sqrt{x}}{x^{2/3}} \right ) ^3$

This is an exercise in the chain rule:
$y^{\prime} = \left [ 3 \left ( \frac{1 + \sqrt{x}}{x^{2/3}} \right )^2 \right ] \cdot \left [ \frac{\frac{1}{2 \sqrt{x}} \cdot x^{2/3} - (1 + \sqrt{x} ) \cdot \frac{2}{3}x^{-1/3} }{x^{4/3}} \right ]$

The first set of [ ] is the derivative of the "outer" function, the cube. The second set of [ ] is the derivative of the fraction inside the ( ). Of course, you still need to simplify this.

-Dan
That's what I had so far, but with all the fractions I got super confused on the simplification.

And for #1, what do I do with the three equations and the three unknowns?

6. Originally Posted by topsquark
So find the derivatives, set them equal to each other, and solve for x.
So wouldn't it be

6x² + 6x5 = 12x5
6x² = 6x5
x²=x5

Now what?

7. Originally Posted by NAPA55
So wouldn't it be

6x² + 6x5 = 12x5
6x² = 6x5
x²=x5

Now what?
$x^5 = x^2$

$x^5 - x^2 = 0$

$x^2(x^3 - 1) = 0$

$x^2(x - 1)(x^2 + x + 1) = 0$

Set each of these factors equal to 0 and you find that
$x = 0, 1$
(The third term has only complex zeros, so we may ignore the solutions from this factor.)

-Dan