At what ##x## value is the tangent equally inclined to the given curve?

• chwala
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
differentiation.
I had to look this up; will need to read on it.

from my research,

...
I have noted that at equally inclined; the slope value is ##1##.

##\dfrac{dy}{dx} = 2x^2+x=1##

##2x^2+x-1=0##

##x=-1## or ##x=0.5##

the steps are clear; but i need to understand the concept...i guess more reading on my part. ...just sharing in the event one has insight to offer. Cheers.

chwala said:
Homework Statement: See attached.
Relevant Equations: differentiation.

I had to look this up; will need to read on it.

View attachment 342013

from my research,

...
I have noted that at equally inclined; the slope value is ##1##.

##\dfrac{dy}{dx} = 2x^2+x=1##

##2x^2+x-1=0##

##x=-1## or ##x=0.5##

the steps are clear; but i need to understand the concept...i guess more reading on my part. ...just sharing in the event one has insight to offer. Cheers.
It's pretty straightforward. If you graph ##y = \frac 2 3 x^3 + \frac 1 2 x^2##, there are points in the first and third quadrants at which the slope is 1, namely, at (-1, -1/6) and (1/2, 5/24).

chwala
Noted, in general can we also have equally inclined when slope ##=-1##?

chwala said:
Noted, in general can we also have equally inclined when slope ##=-1##?

I would say yes.

chwala
Just for reference, here is the plot of the function and the line ##y = x##.

Pretty suggestive that at x = -1 and 0.5 are the slope is indeed the same slope as ##y = x##.

chwala
chwala said:
Noted, in general can we also have equally inclined when slope ##=-1##?
In general, yes. For this particular case, it never happens as
$$2x^2 + x + 1 = 0$$
has roots ##-1/4 \pm \sqrt{1/16 - 1/2} = -1/4 \pm i \sqrt{7}/4##, which both have non-zero imaginary part.

chwala

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