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Find the equation for the tangent line to the curve y=f(x) at the given x -value. f(x)=e^{x^{3}-1} { at } x=1

User Xeberdee
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1 Answer

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Final answer:

To find the tangent line to the curve at x=1 for the function f(x)=e^{x^3-1}, calculate the derivative, evaluate it at x=1 to get the slope, and then use the point-slope form. The equation of the tangent line at x=1 is y = 3x - 2.

Step-by-step explanation:

To find the equation of the tangent line to the curve y=f(x) at a specific value of x, we need to compute the derivative of the function to find the slope of the tangent at that point. For the function f(x) = e^{x^3-1}, we find the derivative, evaluate it at x = 1, and then use the point-slope form of the line to write the equation of the tangent.

Let's calculate the derivative:

f'(x) = d/dx (e^{x^3-1})

= e^{x^3-1} × d/dx (x^3 - 1) by the chain rule

= e^{x^3-1} × 3x^2

Now, evaluating the derivative at x = 1:

f'(1) = e^{1^3-1} × 3 × 1^2

= e^0 × 3

= 3

The slope of the tangent line at x = 1 is 3, and since we are given the x-value, the point on the curve is (1, f(1)).

f(1) = e^{1^3 - 1}

= e^0

= 1

So the point is (1, 1). Using the point-slope form, y - y1 = m(x - x1), we get:

y - 1 = 3(x - 1)

Thus, the equation of the tangent line at x = 1 is y = 3x - 2.

User Brewmanz
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