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How do you find the slope of a tangent line to the graph of the function #f(x)=x^3-2x+1# at (-1,2)?

Answer 1

Ezekiel Alexander

#y'=1# at #(-1,2)#

First, we have to differentiate the function, #y=x^3-2x+1#

the rule for derivatives is that if. #y=x^n# #dy/dx=nx^(n-1)#

so for,

#y=x^3-2x+1#

#dy/dx=3x^(3-1)-2x^0 +0#

#dy/dx=3x^(2)-2#

so to find the gradient of the tangent we sub in the x value of the point.

#dy/dx=3(-1)^2-2#

#dy/dx=3-2#

#dy/dx=1#

The gradient of the tangent is 1 at the point (-1,2)

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Answer 2

Isabella Ackerman

To find the slope of a tangent line to the graph of a function at a specific point, you can use the derivative of the function. The derivative of f(x) is f'(x), which represents the rate of change of the function at any given point. To find the slope of the tangent line at (-1,2), you need to evaluate the derivative at x = -1.

The derivative of f(x) = x^3 - 2x + 1 is f'(x) = 3x^2 - 2.

Substituting x = -1 into the derivative, we get f'(-1) = 3(-1)^2 - 2 = 3 - 2 = 1.

Therefore, the slope of the tangent line to the graph of f(x) = x^3 - 2x + 1 at (-1,2) is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.