# What is the equation of the line tangent to #f(x)= x^5-e^x # at #x=2#?

exactly or

approximately.

You need to take the derivative and the evaluate it at x=2.

We have

We take

using the Linearity of differentiation.

Recall the equation of the line,

so we have

Rearranging we get

Putting everything back into the line we have,

exactly or

approximately.

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The equation of the line tangent to f(x) = x^5 - e^x at x = 2 is y = 32x - e^2 - 62.

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To find the equation of the tangent line to the function ( f(x) = x^5 - e^x ) at ( x = 2 ), follow these steps:

- Find the slope of the tangent line by taking the derivative of the function ( f(x) ) with respect to ( x ) and evaluating it at ( x = 2 ).

[ f'(x) = 5x^4 - e^x ]

- Evaluate ( f'(2) ) to find the slope of the tangent line.

[ f'(2) = 5(2)^4 - e^2 ]

- Substitute ( x = 2 ) into the original function ( f(x) ) to find the ( y )-coordinate of the point where the tangent line touches the curve.

[ f(2) = (2)^5 - e^2 ]

- Use the point-slope form of the equation of a line to write the equation of the tangent line.

[ y - f(2) = f'(2)(x - 2) ]

- Simplify the equation to obtain the final equation of the tangent line.

[ y - [(2)^5 - e^2] = [5(2)^4 - e^2](x - 2) ]

[ y - [32 - e^2] = [80 - e^2](x - 2) ]

[ y = (80 - e^2)x - 128 + e^2 ]

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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.

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.

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.

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