# Find the x-intercept of the tangent line to the graph of f at the point P(0, f(0)) when f(x)=e^x(4cosx-5sinx)

You need to find first the equation of the tangent line to to the given curve and then you may evaluate x intercept of the tangent line.

You should remember the equation that gives the tangent line to the curve at a point such that:

`y - f(x_0) = f'(x_0)(x - x_0)`

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You need to find first the equation of the tangent line to to the given curve and then you may evaluate x intercept of the tangent line.

You should remember the equation that gives the tangent line to the curve at a point such that:

`y - f(x_0) = f'(x_0)(x - x_0)`

Hence, you need to find the derivative of the given function at the point `(0,f(0)),`  using the product rule, such that:

`f'(x) = (e^x)'(4cos x-5sin x) + e^x(4cos x-5sin x)'`

`f'(x) = e^x(4cos x-5sin x) + e^x(-4sin x - 5cos x)`

Factoring out `e^x`  yields:

`f'(x) = e^x(4 cos x - 5 sin x - 4 sin x - 5 cos x)`

`f'(x) = e^x(-cos x - 9 sin x)`

You may evaluate `f'(0)`  such that:

`f'(0) = e^0(-cos 0 - 9sin 0) => f'(0) = -1`

You need to write the equation of the tangent line at the point `(0,f(0)) `  such that:

`y - f(0) = f'(0)(x - 0)`

You need to evaluate f(0) such that:

`f(0) = (e^0)(4cos 0-5sin 0) => f(0) = -4`

Hence, substituting the found values in equation of tangent line yields:

`y + 4 = -1*x => y = -x - 4`

Since you know now the equation of tangent line, you may evaluate x intercept considering `y=0`  such that:

`0 = -x - 4 => x = -4`

Hence, evaluating the x intercept of the tangent line to the given curve yields `x = -4` .

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