Is this differentiation right.

question:

(2x-5)^3

My Answer:

6(2x-5)^2

If I do this how do I find the gradient. If I sub x=0 i don't get what I want.

Please help.

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- Feb 9th 2010, 12:54 AMAwsom Guydifferentiation
Is this differentiation right.

question:

(2x-5)^3

My Answer:

6(2x-5)^2

If I do this how do I find the gradient. If I sub x=0 i don't get what I want.

Please help. - Feb 9th 2010, 01:04 AMArchie Meade
The differentiation is correct.

That is the curve gradient for any x.

If you sub in x=0, you get the slope of the tangent to the curve at x=0,

but not the slope of the tangent that passes through the origin if that's what you were looking for.

What were you trying to calculate ? - Feb 9th 2010, 01:05 AMProve It
- Feb 9th 2010, 01:06 AMAwsom Guy
My question says: Find the point where the curve y=(2x-5)^3 cuts the x-axis and find the slope of the tangent at this point.

- Feb 9th 2010, 01:09 AMProve It
Well then do as you are instructed.

The find where the curve cuts the $\displaystyle x$ axis, let $\displaystyle y = 0$.

So $\displaystyle 0 = (2x - 5)^3$

$\displaystyle 2x - 5 = 0$

$\displaystyle x = \frac{5}{2}$.

So you are not evaluating the derivative at $\displaystyle x = 0$. You are evaluating the derivative at $\displaystyle x = \frac{5}{2}$. - Feb 9th 2010, 01:12 AMAwsom Guy
do i then place it in the y=mx+b formula to calculate the slope at the tagent at the point.

- Feb 9th 2010, 01:14 AMAwsom Guy
just so everyone knows the answer is m=0

- Feb 9th 2010, 01:22 AMArchie Meade
no need,

the derivative__is__the slope or gradient of the tangent.

That's zero when y=0 also in this case. - Feb 9th 2010, 01:24 AMAwsom Guy
But the derivative isn't 0 it is 6(2x-5)^2

- Feb 9th 2010, 01:27 AMArchie Meade
It is when $\displaystyle x=\frac{5}{2}$ or y=0 !

- Feb 9th 2010, 01:29 AMAwsom Guy
Thank you so much for helping me.

- Feb 9th 2010, 01:32 AMArchie Meade
here's the graph,

the tangent is the x axis itself at x=2.5. - Feb 9th 2010, 10:55 PMAwsom Guy
thanks