# 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)

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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` .

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

Given

## f(x)=e^x(4cosx-5sinx)

=> f(0) = e^0

= 0

So slope of the tangent line at (0,1) will be,

m = f'(x=0)

To find out f'(x) I proceed as follows:

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

=> ln[ f(x)] = x * [4 Cos(x) - 5 Sin(x) ]

= > (d/dx) ln[ f(x)]  = (d/dx) {x * [4 Cos(x) - 5 Sin(x) ]}

=> [1/f(x)] f'(x) = [4 Cos(x) - 5 Sin(x) ]

+ x* (- 4 Sin(x) - 5 Cos(x) )

=> f'(x) = f(x) * { [4 Cos(x) - 5 Sin(x) ]

+ x* (- 4 Sin(x) - 5 Cos(x) ) }

at x = 0, f(x) = 1

So

f'(0) = 1 * { [4 Cos(0) - 5 Sin(0) ]
+ 0* (- 4 Sin(0) - 5 Cos(0) ) }

= 4

Hence slope of the tanget at (0,f(0)) is m = 4

So the equation of the line is:

y = 4 * x + C

as the tanget touches the graph at x = 0 and f(0) = y = 1  so

1 = 4*0 + C

=> c = 1

So the equation of the tangent line is:

y = 4*x + 1

X intercept is the x-value where y becomes 0

So the x-intercept is:

0 = 4*x + 1

=> x = - 1/4