Kazane said:Homework Statement
f(x)=x1/3
use the definition of f'(a) to calculate the slope of the curve at x=1
(Hint: by rationalizing the numerator. useful formula a3 -b3=(a-b)(a2+ab+b2)
The Attempt at a Solution
f'(x)=lim (x->1) f(x)-f(1)/x-1
=lim (x->1) x1/3-11/3/ x-1
How can I do x1/3-11/3 to a3 -b3=(a-b)(a2+ab+b2)?
I'm confused so please help!
Let [itex]a= x^{1/3}[/itex], [itex]b= y^{1/3}[/itex]Kazane said:Homework Statement
f(x)=x1/3
use the definition of f'(a) to calculate the slope of the curve at x=1
(Hint: by rationalizing the numerator. useful formula a3 -b3=(a-b)(a2+ab+b2)
The Attempt at a Solution
f'(x)=lim (x->1) f(x)-f(1)/x-1
=lim (x->1) x1/3-11/3/ x-1
How can I do x1/3-11/3 to a3 -b3=(a-b)(a2+ab+b2)?
I'm confused so please help!
The slope of a curve at a certain point is defined as the rate of change of the curve at that specific point. In other words, it measures how steep or flat the curve is at that point.
To calculate the slope of a curve at a specific point, we use the derivative function. At x=1, we plug in the value of 1 into the derivative function and solve for the slope.
The slope of the curve at x=1 tells us the rate of change of the curve at that point. It helps us understand how fast the curve is increasing or decreasing at x=1.
The slope of the curve at x=1 is the slope of the tangent line at that point on the curve. The tangent line is a straight line that touches the curve at that point and represents the slope of the curve at that point.
Calculating the slope of a curve at x=1 is important because it helps us understand the behavior of the curve at that specific point. It also allows us to make predictions about the curve and analyze its overall shape and trends.