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Mathematics 21 Online
OpenStudy (anonymous):

at what points does the curve y= x^3 slope at 45 degrees?

OpenStudy (anonymous):

slope at 45 degrees with the x-axis, right?

OpenStudy (anonymous):

yes i think so

OpenStudy (anonymous):

the line tangent to the curve must have a slope of 1/2-->the derivative of the x^3=1/2

OpenStudy (anonymous):

the derivative of the line tangent to the curve at point a is 1

OpenStudy (anonymous):

i don't get it

OpenStudy (anonymous):

is this for calculus?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

well, think about it. The slope of line y=x is 45 degrees, and it has a slope of 1

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so the derivative of the curve at point a is the slope of the line tangent to the curve at point a = 1

OpenStudy (anonymous):

meaning the derivative of x^3 at a is 1

OpenStudy (anonymous):

OHHHHH i understand now thanks ^^

OpenStudy (anonymous):

so do u need the solution, or you got it?

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