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A function f has domain R and range {y ∈ R : y ≥ e} The graph of y = f(x) is shown - AQA - A-Level Maths: Pure - Question 7 - 2018 - Paper 2
Question 7
A function f has domain R and range {y ∈ R : y ≥ e} The graph of y = f(x) is shown. The gradient of the curve at the point (x, y) is given by dy/dx = (x - 1)e^x ... show full transcript
Worked Solution & Example Answer:A function f has domain R and range {y ∈ R : y ≥ e} The graph of y = f(x) is shown - AQA - A-Level Maths: Pure - Question 7 - 2018 - Paper 2
Step 1
Find an expression for f(x) using integration
Answer
To find the expression for f(x), we first integrate the given derivative:
rac{dy}{dx} = (x - 1)e^xWe can use integration by parts, where we let:
- u = (x - 1)
- dv = e^x dx
This gives us:
- du = dx
- v = e^x
Now applying the integration by parts formula:
egin{align*} ext{Integral} & = uv - ext{Integral}(v du) \\ & = (x - 1)e^x - ext{Integral}(e^x dx) \\ & = (x - 1)e^x - e^x + C \\ & = xe^x - 2e^x + C ext{where C is the constant of integration.} \end{align*}Thus, we have:
egin{align*} f(x) & = xe^x - 2e^x + C\end{align*}Step 2
Justify minimum value y = e
Answer
Next, we must determine the minimum value of y. From the range of the function, we know that:
To find when this occurs, we set the derivative equal to 0:
rac{dy}{dx} = 0 ext{ at } (x - 1)e^x = 0.This leads to the conclusion that x must equal 1, hence the minimum value occurs at:
For the curve to pass through the point (1, e), substituting gives:
which implies:
Step 3