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CBSE Sample Paper 2023-24 · Section B · 2 marks

Q22: Interval where f(x)=xexf(x) = xe^x is increasing

Sample 2023-24Ch 6 · Application of DerivativesCh 5 · Continuity and Differentiability
The question

Find the interval/s in which the function f:R→Rf: \mathbb{R} \to \mathbb{R} defined by f(x)=xexf(x) = xe^x⁠, is increasing.

Exam answer
f(x)=xex⇒f′(x)=ex(x+1)f(x) = xe^x \Rightarrow f'(x) = e^x(x + 1)
Why?

xexxe^x is a product, so use the product rule:

ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

With u=xu = x and v=exv = e^x⁠:

u=x,dudx=1v=ex,dvdx=ex\begin{aligned} u &= x, & \frac{du}{dx} &= 1 \\ v &= e^x, & \frac{dv}{dx} &= e^x \end{aligned} f′(x)=x⋅ex+ex⋅1=xex+ex=ex(x+1)(take out ex)\begin{aligned} f'(x) &= x\cdot e^x + e^x\cdot 1 \\ &= xe^x + e^x \\ &= e^x(x + 1) \quad (\text{take out } e^x) \end{aligned}

For f(x)f(x) to be increasing, we have

f′(x)=ex(x+1)≥0f'(x) = e^x(x + 1) \ge 0
Why?

The sign of the derivative tells us the direction of the graph. Where f′(x)≥0f'(x) \ge 0 (and f′f' is zero only at single points), ff is increasing.

As ex>0e^x > 0 for all x∈Rx \in \mathbb{R}⁠,

x+1≥0⇒x≥−1x + 1 \ge 0 \Rightarrow x \ge -1
Why?

exe^x is positive for every real xx⁠, so dividing by it does not change the inequality. The sign of f′(x)f'(x) is just the sign of x+1x + 1⁠.

ex(x+1)≥0x+1≥0(divide by ex>0)x≥−1(subtract 1)\begin{aligned} e^x(x + 1) &\ge 0 \\ x + 1 &\ge 0 \quad (\text{divide by } e^x > 0) \\ x &\ge -1 \quad (\text{subtract } 1) \end{aligned}

The end point is included: f′(−1)=0f'(-1) = 0 only at that one point, so ff is still increasing on [−1,∞)[-1, \infty)⁠.

Hence, the required interval where f(x)f(x) increases is [−1,∞)[-1, \infty)⁠.

Where this comes from

Chapter 6, Application of Derivatives: Increasing functions: sign of the derivative.

Question type: Find where a function increases and decreases

How to spot it: click to see

Type 5. Find where a function increases and decreases

Spot it: "Find the intervals in which ff is increasing or decreasing", or "strictly increasing or strictly decreasing".

Here: f′(x)=ex(x+1)f'(x) = e^x(x + 1)⁠, and exe^x is always positive, so the sign of f′f' is the sign of x+1x + 1⁠. It is 0 at x=−1x = -1 and positive to the right, giving [−1,∞)[-1, \infty)⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 5, Continuity and Differentiability.

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