Question Find the largest possible value of $f(15)$ if $f(x)$ is continuous and differentiable on $[6,15]$ , $f(6)=-2$ , and $f'(x) \leq 10$ for all $x \in [6,15]$ .

Studdy Solution

STEP 1

Assumptions
1. The function $f(x)$ is continuous on the interval $[6,15]$ .
2. The function $f(x)$ is differentiable on the interval $[6,15]$ .
3. The value of $f(6)$ is given as $-2$ .
4. The derivative of the function $f'(x)$ is less than or equal to $10$ for all $x$ in the interval $[6,15]$ .

STEP 2

We will use the Mean Value Theorem (MVT) which states that if a function is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , then there exists at least one point $c$ in $(a, b)$ such that:
$f'(c) = \frac{f(b) - f(a)}{b - a}$

STEP 3

Since $f(x)$ is continuous and differentiable on $[6,15]$ and $f'(x) \leq 10$ for all $x$ in $[6,15]$ , we can apply the MVT to find an upper bound for $f(15) - f(6)$ .

STEP 4

Using the MVT, we know there exists some $c \in (6,15)$ such that:
$f'(c) = \frac{f(15) - f(6)}{15 - 6}$

STEP 5

Because $f'(x) \leq 10$ for all $x \in [6,15]$ , we have:
$f'(c) \leq 10$

STEP 6

Substitute the inequality for $f'(c)$ into the equation from the MVT:
$10 \geq \frac{f(15) - f(6)}{15 - 6}$

STEP 7

Multiply both sides of the inequality by $15 - 6$ to isolate $f(15) - f(6)$ on one side:
$(15 - 6) \cdot 10 \geq f(15) - f(6)$

STEP 8

Calculate the left side of the inequality:
$9 \cdot 10 \geq f(15) - f(6)$
$90 \geq f(15) - f(6)$

STEP 9

Add $f(6)$ to both sides of the inequality to solve for $f(15)$ :
$90 + f(6) \geq f(15)$

STEP 10

Substitute the given value of $f(6) = -2$ into the inequality:
$90 + (-2) \geq f(15)$
$88 \geq f(15)$

STEP 11

The largest possible value of $f(15)$ , given the constraints, is $88$ .
Therefore, the correct answer is $88$ .

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Question Find the largest possible value of $f(15)$ if $f(x)$ is continuous and differentiable on $[6,15]$ , $f(6)=-2$ , and $f'(x) \leq 10$ for all $x \in [6,15]$ .

Studdy Solution

STEP 1

Assumptions
1. The function $f(x)$ is continuous on the interval $[6,15]$ .
2. The function $f(x)$ is differentiable on the interval $[6,15]$ .
3. The value of $f(6)$ is given as $-2$ .
4. The derivative of the function $f'(x)$ is less than or equal to $10$ for all $x$ in the interval $[6,15]$ .

STEP 2

We will use the Mean Value Theorem (MVT) which states that if a function is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , then there exists at least one point $c$ in $(a, b)$ such that:
$f'(c) = \frac{f(b) - f(a)}{b - a}$

STEP 3

Since $f(x)$ is continuous and differentiable on $[6,15]$ and $f'(x) \leq 10$ for all $x$ in $[6,15]$ , we can apply the MVT to find an upper bound for $f(15) - f(6)$ .

STEP 4

Using the MVT, we know there exists some $c \in (6,15)$ such that:
$f'(c) = \frac{f(15) - f(6)}{15 - 6}$

STEP 5

Because $f'(x) \leq 10$ for all $x \in [6,15]$ , we have:
$f'(c) \leq 10$

STEP 6

Substitute the inequality for $f'(c)$ into the equation from the MVT:
$10 \geq \frac{f(15) - f(6)}{15 - 6}$

STEP 7

Multiply both sides of the inequality by $15 - 6$ to isolate $f(15) - f(6)$ on one side:
$(15 - 6) \cdot 10 \geq f(15) - f(6)$

STEP 8

Calculate the left side of the inequality:
$9 \cdot 10 \geq f(15) - f(6)$
$90 \geq f(15) - f(6)$

STEP 9

Add $f(6)$ to both sides of the inequality to solve for $f(15)$ :
$90 + f(6) \geq f(15)$

STEP 10

Substitute the given value of $f(6) = -2$ into the inequality:
$90 + (-2) \geq f(15)$
$88 \geq f(15)$

STEP 11

The largest possible value of $f(15)$ , given the constraints, is $88$ .
Therefore, the correct answer is $88$ .

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Question Find the largest possible value of f(15)f(15)f(15) if f(x)f(x)f(x) is continuous and differentiable on [6,15][6,15][6,15], f(6)=−2f(6)=-2f(6)=−2, and f′(x)≤10f'(x) \leq 10f′(x)≤10 for all x∈[6,15]x \in [6,15]x∈[6,15].

Studdy Solution

STEP 1

STEP 2

STEP 3

Since f(x)f(x)f(x) is continuous and differentiable on [6,15][6,15][6,15] and f′(x)≤10f'(x) \leq 10f′(x)≤10 for all xxx in [6,15][6,15][6,15], we can apply the MVT to find an upper bound for f(15)−f(6)f(15) - f(6)f(15)−f(6).

STEP 4

Using the MVT, we know there exists some c∈(6,15)c \in (6,15)c∈(6,15) such that:f′(c)=f(15)−f(6)15−6f'(c) = \frac{f(15) - f(6)}{15 - 6}f′(c)=15−6f(15)−f(6)​

STEP 5

Because f′(x)≤10f'(x) \leq 10f′(x)≤10 for all x∈[6,15]x \in [6,15]x∈[6,15], we have:f′(c)≤10f'(c) \leq 10f′(c)≤10

STEP 6

Substitute the inequality for f′(c)f'(c)f′(c) into the equation from the MVT:10≥f(15)−f(6)15−610 \geq \frac{f(15) - f(6)}{15 - 6}10≥15−6f(15)−f(6)​

STEP 7

Multiply both sides of the inequality by 15−615 - 615−6 to isolate f(15)−f(6)f(15) - f(6)f(15)−f(6) on one side:(15−6)⋅10≥f(15)−f(6)(15 - 6) \cdot 10 \geq f(15) - f(6)(15−6)⋅10≥f(15)−f(6)

STEP 8

Calculate the left side of the inequality:9⋅10≥f(15)−f(6)9 \cdot 10 \geq f(15) - f(6)9⋅10≥f(15)−f(6)90≥f(15)−f(6)90 \geq f(15) - f(6)90≥f(15)−f(6)

STEP 9

Add f(6)f(6)f(6) to both sides of the inequality to solve for f(15)f(15)f(15):90+f(6)≥f(15)90 + f(6) \geq f(15)90+f(6)≥f(15)

STEP 10

Substitute the given value of f(6)=−2f(6) = -2f(6)=−2 into the inequality:90+(−2)≥f(15)90 + (-2) \geq f(15)90+(−2)≥f(15)88≥f(15)88 \geq f(15)88≥f(15)

STEP 11

The largest possible value of f(15)f(15)f(15), given the constraints, is 888888.Therefore, the correct answer is 888888.

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Question Find the largest possible value of $f(15)$ if $f(x)$ is continuous and differentiable on $[6,15]$ , $f(6)=-2$ , and $f'(x) \leq 10$ for all $x \in [6,15]$ .

Since $f(x)$ is continuous and differentiable on $[6,15]$ and $f'(x) \leq 10$ for all $x$ in $[6,15]$ , we can apply the MVT to find an upper bound for $f(15) - f(6)$ .

Using the MVT, we know there exists some $c \in (6,15)$ such that:
$f'(c) = \frac{f(15) - f(6)}{15 - 6}$

Because $f'(x) \leq 10$ for all $x \in [6,15]$ , we have:
$f'(c) \leq 10$

Substitute the inequality for $f'(c)$ into the equation from the MVT:
$10 \geq \frac{f(15) - f(6)}{15 - 6}$

Multiply both sides of the inequality by $15 - 6$ to isolate $f(15) - f(6)$ on one side:
$(15 - 6) \cdot 10 \geq f(15) - f(6)$

Calculate the left side of the inequality:
$9 \cdot 10 \geq f(15) - f(6)$
$90 \geq f(15) - f(6)$

Add $f(6)$ to both sides of the inequality to solve for $f(15)$ :
$90 + f(6) \geq f(15)$

Substitute the given value of $f(6) = -2$ into the inequality:
$90 + (-2) \geq f(15)$
$88 \geq f(15)$

The largest possible value of $f(15)$ , given the constraints, is $88$ .
Therefore, the correct answer is $88$ .