A function $f$ has domain $ ext{R}$ and range $ ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{ extgreater} ext{e}$. The graph of $y = f(x)$ is shown. The gradient of the curve at the point $(x,y)$ is given by $rac{dy}{dx} = (x-1)e^{x}$. Find an expression for $f(x)$. Fully justify your answer.

A-Level AQA Maths Mechanics Kinematics Graphs: A function $f$ has domain $ ext{

A function $f$ has domain $ ext{R}$ and range $ ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{ extgreater} ext{e}$ - AQA - A-Level Maths Mechanics - Question 7 - 2018 - Paper 2

Question 7

A function $f$ has domain $ ext{R}$ and range $ ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{ extgreater} ext... show full transcript

Worked Solution & Example Answer:A function $f$ has domain $ ext{R}$ and range $ ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{ extgreater} ext{e}$ - AQA - A-Level Maths Mechanics - Question 7 - 2018 - Paper 2

Step 1

Integrate using integration by parts

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Answer

To find $f(x)$ , we start from the gradient expression: $\frac{dy}{dx} = (x-1)e^{x}.$
Using integration by parts, let:

$u = (x-1)$ , then $du = dx$ , and
$dv = e^{x}dx$ , then $v = e^{x}$ .

The integration by parts formula is: $\int u \, dv = uv - \int v \, du.$

Substituting in, we get: $f(x) = (x-1)e^{x} - \int e^{x} \, dx = (x-1)e^{x} - e^{x} + C = xe^{x} - 2e^{x} + C,$
where $C$ is the constant of integration.

Step 2

Justify the minimum y value

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Answer

Next, we need to use the information about the minimum value of the function. From the expression for $rac{dy}{dx}$ , we set: $\frac{dy}{dx} = 0 \Rightarrow (x-1)e^{x} = 0.$
Since $e^{x}$ is never zero, we find: $x - 1 = 0 \Rightarrow x = 1.$
Now substituting $x = 1$ into our expression for $f(x)$ gives: $f(1) = (1)e^{1} - 2e^{1} + C = e - 2e + C = -e + C.$

The range of $f(x)$ specifies that the minimum value is at least $e$ , thus: $-e + C = e \Rightarrow C = 2e.$

Step 3

Final expression for f(x)

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Answer

Finally, substituting the value of $C$ back into our expression for $f(x)$ gives: $f(x) = xe^{x} - 2e^{x} + 2e,$ completing the solution.

A function $f$ has domain $ ext{R}$ and range $ ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{ extgreater} ext{e}$ - AQA - A-Level Maths Mechanics - Question 7 - 2018 - Paper 2

Worked Solution & Example Answer:A function $f$ has domain $ ext{R}$ and range $ ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{y} ext{ } ext{ } ext{ } ext{ extgreater} ext{e}$ - AQA - A-Level Maths Mechanics - Question 7 - 2018 - Paper 2

Integrate using integration by parts

Justify the minimum y value

Final expression for f(x)

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