Exponential and Logarithmic Graphs | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Inverse graphs swap the coordinates

A function’s domain is its allowed inputs; its range is its possible outputs. Since $e^x$ takes any real input and gives a positive output, $\ln x$ takes positive inputs and gives any real output.

If $(u,v)$ lies on $y=e^x$ , then $v=e^u$ and $u=\ln v$ . Thus $(v,u)$ lies on $y=\ln x$ . Swapping coordinates reflects a graph in $y=x$ .

Swapping (0,1) and (1,0) reflects the exponential and logarithm in y = x.

An asymptote is a straight line the curve approaches arbitrarily closely in a stated direction. Here $e^x$ approaches $y=0$ as $x$ decreases without limit. It never reaches zero. The inverse curve approaches $x=0$ from the right as its output decreases without limit.

Your turnPractice

Sketch $y=e^x$ and $y=\ln x$ . Label their axis intercepts and asymptotes. What point on the second corresponds to $(-1,1/e)$ on the first?

Show worked answer

Answer

(1/e,-1)

The exponential crosses the vertical axis at (0,1), with asymptote $y=0$ . The log crosses the horizontal axis at (1,0), with asymptote $x=0$ . Neither crosses its asymptote.

The sign of k chooses growth or decay

For $y=e^{kx}$ , increasing $x$ increases the exponent if $k>0$ , but decreases it if $k<0$ . Every curve passes through (0,1), since $e^{k(0)}=1$ .

Predict the graph when $k=0$ , then move the control across zero. Notice which end approaches $y=0$ .

Exponent coefficient k = 1

Growth: y approaches 0 to the left.

The green point (0,1) stays fixed as k changes. The blue point shows the output at x = 1.

For any $k\ne0$ , the outputs are positive and the horizontal asymptote is $y=0$ . For $k=0$ , the function is the constant line $y=1$ : the output never changes, and its range is only 1.

Your turnPractice

Describe $y=e^{-2x}$ : its intercept, direction, range and which end approaches zero. Does a negative exponent make its output negative?

Show worked answer

It decreases through (0,1), has range $y>0$ , and approaches zero as $x$ increases without limit. Its output remains positive: $e^{-2x}=1/e^{2x}$ .

Inverse graphs swap the coordinates

If $(u,v)$ lies on $y=e^x$ , then $v=e^u$ and $u=\ln v$ . Thus $(v,u)$ lies on $y=\ln x$ . Swapping coordinates reflects a graph in $y=x$ .

Swapping (0,1) and (1,0) reflects the exponential and logarithm in y = x.

Your turnPractice

Sketch $y=e^x$ and $y=\ln x$ . Label their axis intercepts and asymptotes. What point on the second corresponds to $(-1,1/e)$ on the first?

Show worked answer

Answer

(1/e,-1)

The exponential crosses the vertical axis at (0,1), with asymptote $y=0$ . The log crosses the horizontal axis at (1,0), with asymptote $x=0$ . Neither crosses its asymptote.

The sign of k chooses growth or decay

For $y=e^{kx}$ , increasing $x$ increases the exponent if $k>0$ , but decreases it if $k<0$ . Every curve passes through (0,1), since $e^{k(0)}=1$ .

Predict the graph when $k=0$ , then move the control across zero. Notice which end approaches $y=0$ .

Exponent coefficient k = 1

Growth: y approaches 0 to the left.

The green point (0,1) stays fixed as k changes. The blue point shows the output at x = 1.

For any $k\ne0$ , the outputs are positive and the horizontal asymptote is $y=0$ . For $k=0$ , the function is the constant line $y=1$ : the output never changes, and its range is only 1.

Your turnPractice

Describe $y=e^{-2x}$ : its intercept, direction, range and which end approaches zero. Does a negative exponent make its output negative?

Show worked answer

It decreases through (0,1), has range $y>0$ , and approaches zero as $x$ increases without limit. Its output remains positive: $e^{-2x}=1/e^{2x}$ .