Solving for Tangent Lines and Range of Slopes for a Given Curve

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To find the equation of the tangent line to the curve y = x^3 - 4x + 1 at the point (2,1), first calculate the derivative, y', which gives the slope at that point. The derivative is y' = 3x^2 - 4, so at x = 2, the slope is 8. For part (b), to determine the range of slopes, analyze the derivative function, which is a quadratic that opens upwards, indicating that the slope will vary from negative infinity to positive infinity. In the second question, the derivative of y = x - 3√x can be found using the power rule, yielding y' = 1 - (1/3)x^(-2/3). Understanding these derivatives is essential for solving both problems effectively.
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Hi all





I hve two Q I want the explaine how to solve



Q1 :(A) Find an equation for tangent to curve y = X^3 - 4X + 1 at the point (2,1)



(b ) What is the range of values of the curve's slope





Number ( A ) I can solve it but ( B) I face problem to solve







Q 2 derivative y = x - 3root X



please I want the explaine how to solve How to solve each one .
 
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Use the power rule for Q2. y = x - x^(1/3)
y' = 1 - x^(-2/3)/3
Power rule is d/dx x^n = nx^(n-1), which I'm sure you know..
 
r-soy said:
Hi all

I hve two Q I want the explaine how to solve
Q1 :(A) Find an equation for tangent to curve y = X^3 - 4X + 1 at the point (2,1)
(b ) What is the range of values of the curve's slope

Number ( A ) I can solve it but ( B) I face problem to solve

Q 2 derivative y = x - 3root X

please I want the explaine how to solve How to solve each one .
In problem 1, what did you get for y'? You need that function so that you can find the range of values of the slopes of the tangent lines for the curve.
 
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