How to evaluate function notation at special inputs

Key concept: These exercises test whether you can substitute correctly and keep algebraic signs under control. The main skill is careful replacement, especially when the input is negative or symbolic.

Step by step

27. $f(x)=2x-5$

Substitute the requested inputs:

• $f(-3)=2(-3)-5=-6-5=\boxed{-11}$
• $f(2)=2(2)-5=4-5=\boxed{-1}$
• $f(-a)=2(-a)-5=\boxed{-2a-5}$
• $-f(a)=-\bigl(2a-5\bigr)=\boxed{-2a+5}$
• $f(a+h)=2(a+h)-5=\boxed{2a+2h-5}$

28. $f(x)=-5x^2+2x-1$

• $f(-3)=-5(9)+2(-3)-1=-45-6-1=\boxed{-52}$
• $f(2)=-5(4)+4-1=-20+4-1=\boxed{-17}$
• $f(-a)=-5a^2-2a-1=\boxed{-5a^2-2a-1}$
• $-f(a)=-(-5a^2+2a-1)=\boxed{5a^2-2a+1}$
• $f(a+h)=-5(a+h)^2+2(a+h)-1$

Expand:

$-5(a^2+2ah+h^2)+2a+2h-1$

$\boxed{-5a^2-10ah-5h^2+2a+2h-1}$

30. $f(x)=\frac{6x-1}{5x+2}$

• $f(-3)=\frac{-18-1}{-15+2}=\frac{-19}{-13}=\boxed{\frac{19}{13}}$
• $f(2)=\frac{12-1}{10+2}=\boxed{\frac{11}{12}}$
• $f(-a)=\frac{-6a-1}{-5a+2}=\boxed{\frac{6a+1}{5a-2}}$
• $-f(a)=\boxed{-\frac{6a-1}{5a+2}}$
• $f(a+h)=\boxed{\frac{6a+6h-1}{5a+5h+2}}$

31. $f(x)=|x-1|-|x+1|$

• $f(-3)=|-4|-|-2|=4-2=\boxed{2}$
• $f(2)=|1|-|3|=1-3=\boxed{-2}$
• $f(-a)=|-a-1|-|-a+1|=\boxed{|a+1|-|1-a|}$
• $-f(a)=\boxed{-|a-1|+|a+1|}$
• $f(a+h)=\boxed{|a+h-1|-|a+h+1|}$

33. Given $g(x)=x^2+2x$, evaluate $\frac{g(x)-g(a)}{x-a}$, $x\ne a$

First compute:

$g(x)-g(a)=x^2+2x-(a^2+2a)$

Factor:

$x^2-a^2+2(x-a)=(x-a)(x+a)+2(x-a)$

$=(x-a)(x+a+2)$

So,

$\boxed{\frac{g(x)-g(a)}{x-a}=x+a+2}$

34. Given $k(t)=2t-1$

(a) $k(2)=2(2)-1=\boxed{3}$

(b) Solve $k(t)=7$:

$2t-1=7$ $2t=8$ $t=\boxed{4}$

Pitfall alert

Two spots usually trip people up: first, writing $-f(a)$ as $f(-a)$, which is not the same thing; second, expanding $-5(a+h)^2$ and forgetting the minus sign hits every term after the square is expanded.

Try different conditions

If the function changes slightly, the substitution process stays the same. For example, for $f(x)=ax+b$, the values become $f(-3)=-3a+b$ and $f(a+h)=a(a+h)+b$. For rational functions, always check whether the denominator becomes zero at the chosen input.

Further reading

function notation, substitution, difference quotient