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Exponential functions and e^xEdexcel A-Level Maths: Revision notes

Section 1

The function axa^x and its graph

An exponential function has the variable in the power, f(x)=axf(x)=a^x with a>0a>0 and a≠1a\ne1. Its graph:

  • passes through (0,1)(0,1), because a0=1a^0=1;
  • lies above the xx-axis for all xx (since ax>0a^x>0), with the xx-axis as a horizontal asymptote;
  • for a>1a>1 it increases (growth), rising steeply as xx increases; for 0<a<10<a<1 it decreases (decay). The graph of y=(1a)x=a−xy=\left(\frac1a\right)^x=a^{-x} is the reflection of y=axy=a^x in the yy-axis. Example: if a2=25a^2=25 and a>0a>0 then a=5a=5, f(−1)=15f(-1)=\frac15 and f(x)=1125f(x)=\frac1{125} gives x=−3x=-3.
Key termsexponential functionasymptote
Common mistake

Taking a−1a^{-1} as −a-a. A negative power means a reciprocal: 5−1=155^{-1}=\frac15.

Section 2

The function exe^x

The number e≈2.718e\approx2.718 is the special base for which the gradient of y=axy=a^x at (0,1)(0,1) is exactly 11. The function y=exy=e^x has the remarkable property that its gradient at every point equals its yy-value: ddx(ex)=ex\frac{d}{dx}(e^x)=e^x. Its graph passes through (0,1)(0,1), lies above the xx-axis, and increases at an increasing rate. For any other base the gradient is axa^x multiplied by a constant, which is 11 only for a=ea=e.

Key termse

Section 3

The gradient of ekxe^{kx}

For a constant kk, ddx(ekx)=kekx.\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}. The gradient is kk times the function. If k>0k>0 the function grows; if k<0k<0 it decays. Examples: ddx(e3x)=3e3x\frac{d}{dx}(e^{3x})=3e^{3x} and ddx(5e−2x)=−10e−2x\frac{d}{dx}(5e^{-2x})=-10e^{-2x}. For M=80e−0.05tM=80e^{-0.05t}, dMdt=−0.05M\frac{dM}{dt}=-0.05M.

Key termsgradient
Common mistake

Forgetting the factor kk: the derivative of e3xe^{3x} is 3e3x3e^{3x}, not e3xe^{3x}.

Section 4

Why exponential models are used

Since ddx(Aekx)=k×Aekx\frac{d}{dx}(Ae^{kx})=k\times Ae^{kx}, the rate of change is proportional to the current value yy. Many real situations behave like this: population growth (more individuals produce more offspring), radioactive decay (each nucleus decays with a fixed probability), compound interest and cooling. So y=Aekty=Ae^{kt}, with AA the initial value and kk the rate constant, is a suitable model. Limits: unlimited growth is unrealistic (food, space) and a continuous model cannot show whole items.

Key termsexponential model

Section 5

The graph of y=eax+b+cy=e^{ax+b}+c

Start from y=exy=e^x and transform it:

  • eaxe^{ax}: a stretch parallel to the xx-axis, scale factor 1a\frac1a;
  • eax+be^{ax+b}: a horizontal translation (written ea(x+b/a)e^{a(x+b/a)});
  • +c+c: translation cc units up, so the horizontal asymptote is y=cy=c. The yy-intercept is eb+ce^{b}+c and the range is y>cy>c (for aa positive or negative). Example: y=e2x−1+3y=e^{2x-1}+3 has asymptote y=3y=3, yy-intercept 1e+3\frac1e+3 and never meets the xx-axis because y>3y>3.
Key termshorizontal asymptote
Exam tip

To find the intercept, substitute x=0x=0; for the asymptote, ask what the exponential tends to as x→±∞x\to\pm\infty.

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Exam questions on Exponential functions and e^x

  1. A function is given by f(x)=axf(x)=a^x, where a>0a>0, and its graph passes through the point (2, 25)(2,\,25).
    Solve f(x)=1125f(x)=\dfrac{1}{125}.2 marks
  2. Consider the curve y=e2x−1+3y=e^{2x-1}+3.
    Explain why the curve never crosses the xx-axis.2 marks
  3. The mass MM grams of a radioactive sample tt days after it is first measured is modelled by M=80e−0.05tM=80e^{-0.05t}.
    Find dMdt\dfrac{dM}{dt} and hence show that the rate of change of MM is proportional to MM.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).