If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following (a) (f(t) – f(5))/ (t

Home » If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following (a) (f(t) – f(5))/ (t – 5), if t ≠ 5

If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following (a) (f(t) – f(5))/ (t – 5), if t ≠ 5

If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following (a) (f(t) – f(5))/ (t – 5), if t ≠ 5

Solution:

Provided,

f and g are real functions such that $f(x)=x^{2}+7$ and $g(x)=3 x+5$

(a) $(f(t)-f(5)) /(t-5)$ ,

if $t \neq 5$ $f(x)=x^{2}+7$

Substitute $x = t$ in $f(x)$ , we have

$f(t)=t^{2}+7 \ldots \ldots \dots$ (i)

Now, consider the same function,

$f (x) = x^{2} + 7$

Substitute $x = 5$ in $f(x)$ , we have

$f(5)=(5)^{2}+7=25+7=32 \ldots \ldots \dots$ (ii)

From eq.(i) and eq.(ii), we have

$\frac{f(t)-f(5)}{t-5}=\frac{t^{2}+7-32}{t-5}$

$=\frac{t^{2}-25}{t-5}$

$=\frac{t^{2}-5^{2}}{t-5}$

But it is known that $a^{2}-b^{2}=(a+b)(a-b)$ , therefore the above equation becomes, $\frac{f(t)-f(5)}{t-5}=\frac{(t+5)(t-5)}{t-5}$

Now cancel the like terms, we have

$\frac{f(t)-f(5)}{t-5}=t+5$

i

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