(a) Given f(x)= a" where a > 1, f-'(x) = loga(x) (b) In the equation A(t) = Aoekt, if k is positive then as t → 0o, A(t) → %3D 0. (c) It is possible for a polynomial p(x) to have a single complex root.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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True or false

(a) Given f(x) = a" where a > 1, ƒ-'(x) = loga(x)
(b) In the equation A(t) = Aoekt, if k is positive then as t o0, A(t) →
%3D
0.
(c) It is possible for a polynomial p(x) to have a single complex root.
(d) Given the following inequality (x- a)(x - b) > 0, we must have that
x - a >0 or x- b > 0.
(e) Given the graph of a function f(x), the function g(x)=f(-x) is obtained
by flipping the graph of f(x) across the x axis.
(f) A curve in the x y plane defines a function y of x if it passes the
vertical line test.
(g) A curve in the x y plane is one-to-one if it passes the vertical line
test.
(h) For any polynomial f(x) = anx" +x"-1 +
is always all real numbers.
+a1x+a0; the domain
%3D
...
(i) The graphs of functions and their inverses are always symmetric
about the y axis.
(j) The solution to the inequality r- 1 2 is the interval (3, 0).
Transcribed Image Text:(a) Given f(x) = a" where a > 1, ƒ-'(x) = loga(x) (b) In the equation A(t) = Aoekt, if k is positive then as t o0, A(t) → %3D 0. (c) It is possible for a polynomial p(x) to have a single complex root. (d) Given the following inequality (x- a)(x - b) > 0, we must have that x - a >0 or x- b > 0. (e) Given the graph of a function f(x), the function g(x)=f(-x) is obtained by flipping the graph of f(x) across the x axis. (f) A curve in the x y plane defines a function y of x if it passes the vertical line test. (g) A curve in the x y plane is one-to-one if it passes the vertical line test. (h) For any polynomial f(x) = anx" +x"-1 + is always all real numbers. +a1x+a0; the domain %3D ... (i) The graphs of functions and their inverses are always symmetric about the y axis. (j) The solution to the inequality r- 1 2 is the interval (3, 0).
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