Question

T3-EOY-MAT70 10- Choose the correct answer Hint on steps to follow: Find f'(x) or df/dx Then f'(x) Then f(x) (the given integral with upper limit as xo) Then plug in the equation of the tangent: y -f(x) = f'(x)=(x-x) • Simplify it. Given f(x) = Se' dt. Find the equation of the tangent line to f(x) at x = 2 y = e-2x - e? – 1 y=e-2x - 1

Answer 1

Following the give steps lets first find out the function as

f

The given integral is

$f(x)=\int_{0}^{x} e^t dt = \[e^t\]_{0}^{x}=(e^x-e^0)=e^x-1$

so,

f(3) = e" - 1

f(2) = e- 1

f'.) = " + 0 = et

$f'(2)=e^2$

Plug these values in the equation of tangent  $y-f(x_0)=f'(x_0)(x-x_0)$

we have

$y-f(2)=f'(2)(x-2)$

$\implies y-(e^2-1)=e^2(x-2)$

$\implies y-e^2+1=xe^2-2e^2$

$\implies y=xe^2-2e^2+e^2-1=xe^2-e^2-1$

Hence, $y=xe^2-e^2-1$