If $e^{d y / d x} = x + 1$ given that when $x = 0, y = 3$ then minimum (local) value of $y = f (x)$ is

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Solution

$e^{d y / d x}$ = $x + 1$

Taking natural log of both sides,

$\frac{d y}{d x}$ = $l n (x + 1)$ = 0 (for local minima)

$\Rightarrow$ $x + 1 = 1$
$\Rightarrow$ $x = 0$

It is already given that at x = 0 , y = 3 . Thus, at local minima $x = 0$ , $y = f (x) = 3$

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Similar questions

e^{(d y / d x)} = x + 1

x = 0, y = 3

y = f (x)

e^{\frac{d y}{d x}} = x + 1

y = (2) . ln (2) + k

e^{d y / d x} = x + 1

x = 0, y = 3

e^{d y / d x} = x + 1

x = 0, y = 3.

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