Function fusion #1

Calculus Level 3

Let $f(x)=\int_{-1}^{x} e^{t^{2}} dt$ and $h(x) = f(1+g(x))$ where $g(x)$ is defined for all Real $x$ and $g'(x)$ exists for all Real $x$ . It is given that $g(x)<0$ for all $x>0$ . If $h'(1)=e$ and $g'(1)=1$ , then find $g(1)$ .

The answer is -2.

1 solution

Aditya Jain
Sep 30, 2015

Here is a small solution. We have $f(x) = \int_{-1}^{x} e^{t^{2}}dt$
replacing $x$ by $1+g(x)$ , we get this: $f(1+g(x)) = \int_{-1}^{1+g(x)} e^{t^{2}}dt$
Differentiating above using Leibnitz Rule for differentiation of integrals .
$f'(1+g(x))g'(x)=e^{(1+g(x))^{2}}.g'(x)$
$f'(1+g(x))=e^{(1+g(x))^{2}}$ ( result 1 )
Also, $h(x) = f(1+g(x))$
$h'(x) = f'(1+g(x)).g'(x)$ now substituting using result 1
$h'(x) = e^{(1+g(x))^{2}}.g'(x)$
as $x = 1$ we get:
$e = e^{(1+g(1))^{2}}$ as $h(1) = e$
we get: $(1+g(1))^{2} = 1$
$g(1) = 0, -2$
since it is given that $g(x)<0$ for all $x>0$ , we discard 0. Therefore $g(1)=-2$

Function fusion #1

The answer is -2.

1 solution

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