Help with this question please (1 Viewer)

Gandiva · Aug 8, 2011

Chapter: The Derivative
Find any points on these curves where the tangent has gradient -1

y = 1/4x^4 - 1/2x^2 - x

I'm so annoyed right now!

nightweaver066 · Aug 8, 2011

$y = \frac{1}{4x^4} - \frac{1}{2x^2} -x$

First differentiate:
$\frac{dy}{dx} = -\frac{1}{x^5} + \frac{1}{x^3} -1$

For gradient of -1, $\frac{dy}{dx} = -1$
$\therefore -1 = -\frac{1}{x^5} + \frac{1}{x^3} -1$
$-x^5 = -1 + x^2 -x^5$
$x^2 = 1$
$x = \pm 1$

At x = 1, $y = \frac{1}{4} - \frac{1}{2} - 1$
$y = -\frac{5}{4}$
x = -1, $y = \frac{1}{4} - \frac{1}{2} + 1$
$y = \frac{3}{4}$

$\therefore$ The points are: $(1,-\frac{5}{4}),(-1,\frac{3}{4})$

Gandiva · Aug 8, 2011

Hmm are you sure with that? The answer from the textbook is (1, -1and1/4) (-1, 3/4)

SpiralFlex · Aug 8, 2011

Gandiva said:
Hmm are you sure with that? The answer from the textbook is (1, -1*1/4) (-1, 3/4)

$-1 \frac{1}{4}=-\frac{5}{4}$

Hence nightweaver is correct. Unless you mean, $-1$ times $\frac{1}{4}$ .

Gandiva · Aug 9, 2011

Ah thank you guys for the help!

Help with this question please (1 Viewer)

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