Math Problem Statement

Evaluate each function for the given value for problems 14 to 17.

Solution

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Problem 14:

f(x)=x3+2x21andf(3)f(x) = -x^3 + 2x^2 - 1 \quad \text{and} \quad f(-3) Substitute x=3x = -3 into the function: f(3)=(3)3+2(3)21f(-3) = -(-3)^3 + 2(-3)^2 - 1 =(27)+2(9)1=27+181=44= -( -27 ) + 2(9) - 1 = 27 + 18 - 1 = 44

Thus, f(3)=44f(-3) = 44.

Problem 15:

g(x)=13x2andg(1)g(x) = |13 - x^2| \quad \text{and} \quad g(1) Substitute x=1x = 1 into the function: g(1)=1312=131=12=12g(1) = |13 - 1^2| = |13 - 1| = |12| = 12

Thus, g(1)=12g(1) = 12.

Problem 16:

p(x)=x242x+1andp(8)p(x) = \frac{x^2 - 4}{2x + 1} \quad \text{and} \quad p(-8) Substitute x=8x = -8 into the function: p(8)=(8)242(8)+1=64416+1=6015=4p(-8) = \frac{(-8)^2 - 4}{2(-8) + 1} = \frac{64 - 4}{-16 + 1} = \frac{60}{-15} = -4

Thus, p(8)=4p(-8) = -4.

Problem 17:

h(x)=39xandh(2)h(x) = 3 \cdot 9^x \quad \text{and} \quad h(-2) Substitute x=2x = -2 into the function: h(2)=392=3192=3181=381=127h(-2) = 3 \cdot 9^{-2} = 3 \cdot \frac{1}{9^2} = 3 \cdot \frac{1}{81} = \frac{3}{81} = \frac{1}{27}

Thus, h(2)=127h(-2) = \frac{1}{27}.

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Math Problem Analysis

Mathematical Concepts

Polynomials
Absolute Value
Rational Functions
Exponential Functions

Formulas

Polynomial evaluation f(x) = ax^n + bx^(n-1) + ... + c
Absolute value evaluation g(x) = |h(x)|
Rational function evaluation p(x) = (numerator)/(denominator)
Exponential function evaluation h(x) = a * b^x

Theorems

Fundamental of Algebra

Suitable Grade Level

High School