Answer to Question #248929 in Calculus for Sam

Question

The graph of f(x) = ex curves upward in the interval from x = - 1 to x = 1 Interpreting f(x) = ex as the slopes of tangent lines and noting that the larger x is, the larger exis, explain why the graph curves upward. For larger values of x, the graph of f(x) = ex appears to shoot straight up with no curve. Using the tangent line, determine whether this is correct or just an optical illusion.

Answer 1

A graph is said to be concave up (upward) at a point if the tangent line to the graph at that point lies below the graph.

When the function y = f (x) is concave up, the graph of its derivative y = f '(x) is increasing, i.e.

the graph of y = f (x) is concave upward on those intervals where y = f "(x) > 0.

We have $f''(x)=e^x>0$ , so the graph of f(x) = e^x curves upward.

Tangent line for f(x):

$y-y_0=e^{x_0}(x-x_0)$

$g(x)=e^{x_0}(x-x_0)+y_0$

$x=(g(x)-y_0)/e^{x_0}+x_0$

So, for larger values of x:

$x\to x_0$

So, tangent line of f(x) looks like straight vertical line.

For larger values of x, the graph of f(x) = e^x appears to shoot straight up with no curve, and it is not  optical illusion.